Adding SQL query language
This commit is contained in:
Michael Reber
parent
026079a47d
commit
8c9a19ec97
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/*Write a SQL query to calculate the average price of all the products.*/
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SELECT AVG(pro_price) AS "Average Price"
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FROM item_mast;
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/*Write a SQL query to display the average price of each company's products, along with their code.*/
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SELECT AVG(pro_price) AS "Average Price",
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pro_com AS "Company ID"
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FROM item_mast
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OUP BY pro_com;
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/*Write a SQL query to find the number of products with a price more than or equal to Rs.350.*/
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SELECT COUNT(*) AS "Number of Products"
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FROM item_mast
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WHERE pro_price >= 350;
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/*Write a SQL statement find the number of customers who gets at least a gradation for his/her performance.*/
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SELECT COUNT (ALL grade)
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FROM customer;
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/*Write a SQL statement that counts all orders for a date August 17th, 2012.*/
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SELECT COUNT(*)
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FROM orders
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WHERE ord_date='2012-08-17';
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/*Write a SQL statement that counts the number of different non NULL city values for salesmen.*/
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SELECT COUNT(*)
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FROM salesman
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WHERE city IS NOT NULL;
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/*Write a SQL statement to display customer details (ID and purchase amount) whose IDs are within the range 3002 and 3007
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and highest purchase amount is more than 1000.*/
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SELECT customer_id,MAX(purch_amt)
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FROM orders
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WHERE customer_id BETWEEN 3002 and 3007
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GROUP BY customer_id
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HAVING MAX(purch_amt)>1000;
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/*Write a SQL statement to find the average purchase amount of all orders.*/
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SELECT AVG (purch_amt)
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FROM orders;
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/*Write a SQL statement to find the highest purchase amount on a date '2012-08-17' for each salesman with their ID.*/
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SELECT salesman_id,MAX(purch_amt)
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FROM orders
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WHERE ord_date = '2012-08-17'
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GROUP BY salesman_id;
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/*Write a SQL statement to find the highest purchase amount ordered by the each customer on a particular
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date with their ID, order date and highest purchase amount.*/
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SELECT customer_id,ord_date,MAX(purch_amt)
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FROM orders
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GROUP BY customer_id,ord_date;
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/*Write a SQL statement to find the highest purchase amount ordered by the each
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customer with their ID and highest purchase amount. */
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SELECT customer_id,MAX(purch_amt)
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FROM orders
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GROUP BY customer_id;
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/*Write a SQL statement to find the highest purchase amount with their ID and order date, for only those customers
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who have a higher purchase amount in a day is within the list 2000, 3000, 5760 and 6000.*/
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SELECT customer_id,ord_date,MAX(purch_amt)
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FROM orders
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GROUP BY customer_id,ord_date
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HAVING MAX(purch_amt) IN(2000 ,3000,5760, 6000);
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/*Write a SQL statement to find the highest purchase amount with their ID and order date, for only those customers who have
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highest purchase amount in a day is more than 2000.*/
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SELECT customer_id,ord_date,MAX(purch_amt)
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FROM orders
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GROUP BY customer_id,ord_date
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HAVING MAX(purch_amt)>2000.00;
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/*Write a SQL statement to find the highest purchase amount with their ID and order date,
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for those customers who have a higher purchase amount in a day is within the range 2000 and 6000.*/
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SELECT customer_id,ord_date,MAX(purch_amt)
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FROM orders
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GROUP BY customer_id,ord_date
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HAVING MAX(purch_amt) BETWEEN 2000 AND 6000;
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/*Write a SQL statement to find the highest purchase amount with their ID, for only those customers whose ID is within
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the range 3002 and 3007.*/
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SELECT customer_id,MAX(purch_amt)
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FROM orders
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WHERE customer_id BETWEEN 3002 and 3007
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GROUP BY customer_id;
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/*Write a SQL statement to find the highest purchase amount with their ID, for only those salesmen
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whose ID is within the range 5003 and 5008.*/
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SELECT salesman_id,MAX(purch_amt)
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FROM orders
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GROUP BY salesman_id
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HAVING salesman_id BETWEEN 5003 AND 5008;
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/*Write a SQL statement to find the number of salesmen currently listing for all of their customers.*/
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SELECT COUNT (DISTINCT salesman_id)
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FROM orders;
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/*Write a SQL statement to find the total purchase amount for all orders.*/
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SELECT SUM (purch_amt)
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FROM orders;
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/*Write a SQL statement to know how many customers have listed their names.*/
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SELECT COUNT(*)
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FROM customer;
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/*Write a SQL statement to know the maximum purchase amount of all the orders.*/
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SELECT MAX (purch_amt)
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FROM orders;
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/*Write a SQL statement to know the minimum purchase amount of all the orders. */
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SELECT MIN(purch_amt)
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FROM orders;
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/*Write a SQL statement which selects the highest grade for each of the cities of the customers.*/
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SELECT city,MAX(grade)
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FROM customer
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GROUP BY city;
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/*Write a query in SQL to find the number of employees in each department along with the department code.*/
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SELECT emp_dept, COUNT(*)
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FROM emp_details
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GROUP BY emp_dept;
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/*Write a query in SQL to find the sum of the allotment amount of all departments.*/
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SELECT SUM(dpt_allotment)
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FROM emp_department;
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/*Write a query that counts the number of salesmen with their order date and ID registering orders for each day.*/
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SELECT ord_date,salesman_id,COUNT(*)
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FROM orders
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GROUP BY ord_date,salesman_id;
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/*Display all in reverse, where order dates equal to a specified date or
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customer id greater than a specified number and purchase amount less than a specified amount. */
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SELECT *
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FROM orders
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WHERE NOT((ord_date ='2012-08-17'
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OR customer_id>3005)
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AND purch_amt<1000);
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/*Write a SQL query to display all orders where purchase amount less than a specified amount or order date and customer_id must
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not be greater than a specified data and less than a specified ID respectively.*/
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SELECT *
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FROM orders
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WHERE(purch_amt<200 OR
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NOT(ord_date>='2012-02-10'
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AND customer_id<3009));
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/*Write a SQL query to display order number, purchase amount, achived, the unachieved percentage for those order
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which exceeds the 50% of the target value of 6000. */
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SELECT ord_no,purch_amt,
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(100*purch_amt)/6000 AS "Achieved %",
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(100*(6000-purch_amt)/6000) AS "Unachieved %"
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FROM orders
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WHERE (100*purch_amt)/6000>50;
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/*Write a SQL query to display those customers who are neither belongs to the city New York nor grade value is more than 100. */
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SELECT *
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FROM customer
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WHERE NOT (city = 'New York' OR grade>100);
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/*Write a SQL statement to display all customers, who are either belongs to the city New York or had a grade above 100. */
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SELECT *
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FROM customer
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WHERE city = 'New York' OR grade>100;
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/*Write a SQL statement to display all the customers, who are either belongs to the city New York or
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not had a grade above 100. */
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SELECT *
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FROM customer
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WHERE city = 'New York' OR NOT grade>100;
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/*Write a SQL statement to display either those orders which are not issued on date 2012-09-10 and issued by the
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salesman whose ID is 505 and below or those orders which purchase amount is 1000.00 and below.*/
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SELECT *
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FROM orders
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WHERE NOT ((ord_date ='2012-09-10'
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AND salesman_id>505)
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OR purch_amt>1000.00);
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/*Write a SQL statement to display salesman_id, name, city and commission who gets the
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commission within the range more than 0.10% and less than 0.12%.*/
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SELECT salesman_id,name,city,commission
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FROM salesman
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WHERE (commission > 0.10
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AND commission< 0.12);
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/*Write a query in SQL to display all the data of employees that work in department 47 or department 63.*/
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SELECT *
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FROM emp_details
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WHERE emp_dept = 47 OR emp_dept = 63;
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/*Write a query in SQL to find the data of employees whose last name is Dosni or Mardy.*/
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SELECT *
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FROM emp_details
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WHERE emp_lname ='Dosni' OR emp_lname= 'Mardy';
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/*Write a query statement to display all customers in New York who have a grade value above 100.*/
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SELECT *
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FROM customer
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WHERE city = 'New York' AND grade>100;
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/*Write a query to display all customers with a grade above 100.*/
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SELECT *
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FROM customer
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WHERE grade > 100;
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/*Write a SQL statement to arrange the orders according to the order_date in such a manner that
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the latest date will come first then previous dates.*/
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SELECT *
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FROM orders
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ORDER BY ord_date DESC;
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/*Write a SQL statement to display customer name, city and grade in such a manner that,
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the customer holding highest grade will come first.*/
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SELECT cust_name,city,grade
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FROM customer
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ORDER BY 3 DESC;
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/*Write a SQL statement to display the commission with the percent sign ( % ) with salesman ID,
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name and city columns for all the salesmen.*/
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SELECT salesman_id,name,city,'%',commission*100
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FROM salesman;
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/*Write a SQL statement to display the customer name, city, and grade, etc. and the display will
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be arranged according to the smallest customer ID.*/
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SELECT cust_name,city,grade
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FROM customer
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ORDER BY customer_id;
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/*Write a SQL statement to display the orders with all information in such a manner that, the older order
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date will come first and the highest purchase amount of same day will come first.*/
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SELECT *
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FROM orders
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ORDER BY ord_date,purch_amt DESC;
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/*Write a SQL statement to find out the number of orders booked for each day and display it in
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such a format like "For 2001-10-10 there are 15 orders".*/
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SELECT ' For',ord_date,',there are',
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COUNT (DISTINCT ord_no),'orders.'
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FROM orders
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GROUP BY ord_date;
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/*Write a SQL statement to make a report with customer ID in such a manner that, the largest number of orders booked by the
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customer will come first along with their highest purchase amount.*/
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SELECT customer_id, COUNT(DISTINCT ord_no),
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MAX(purch_amt)
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FROM orders
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GROUP BY customer_id
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ORDER BY 2 DESC;
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/*Write a query to display the orders according to the order number arranged by ascending order.*/
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SELECT *
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FROM orders
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ORDER BY ord_no;
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/*1. Write a query in SQL to display the first name, last name, department number, and department name for each employee. */
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SELECT E.first_name , E.last_name ,
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E.department_id , D.department_name
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FROM employees E
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JOIN departments D
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ON E.department_id = D.department_id;
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/*10. Write a query in SQL to display the first name, last name, department number and name, for all employees who have or have
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not any department.*/
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SELECT E.first_name, E.last_name, E.department_id, D.department_name
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FROM employees E
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LEFT OUTER JOIN departments D
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ON E.department_id = D.department_id;
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/*11. Write a query in SQL to display the first name of all employees and the first name of their manager including those who
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does not working under any manager.*/
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SELECT E.first_name AS "Employee Name",
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M.first_name AS "Manager"
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FROM employees E
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LEFT OUTER JOIN employees M
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ON E.manager_id = M.employee_id;
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/*13. Write a query in SQL to display the job title, department name, full name (first and last name ) of employee, and
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starting date for all the jobs which started on or after 1st January, 1993 and ending with on or before 31 August, 1997.*/
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SELECT job_title, department_name, first_name || ' ' || last_name AS Employee_name, start_date
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FROM job_history
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JOIN jobs USING (job_id)
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JOIN departments USING (department_id)
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JOIN employees USING (employee_id)
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WHERE start_date>='1993-01-01' AND start_date<='1997-08-31';
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/*14. Write a query in SQL to display job title, full name (first and last name ) of employee, and the difference between
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maximum salary for the job and salary of the employee.*/
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SELECT job_title, first_name || ' ' || last_name AS Employee_name,
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max_salary-salary AS salary_difference
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FROM employees
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NATURAL JOIN jobs;
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/*15. Write a query in SQL to display the name of the department, average salary and number of employees working in that
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department who got commission.*/
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SELECT department_name, AVG(salary), COUNT(commission_pct)
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FROM departments
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JOIN employees USING (department_id)
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GROUP BY department_name;
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/*17. Write a query in SQL to display the name of the country, city, and the departments which are running there.*/
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SELECT country_name,city, department_name
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FROM countries
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JOIN locations USING (country_id)
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JOIN departments USING (location_id);
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@ -0,0 +1,5 @@
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/*18. Write a query in SQL to display department name and the full name (first and last name) of the manager.*/
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SELECT department_name, first_name || ' ' || last_name AS name_of_manager
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FROM departments D
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JOIN employees E
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ON (D.manager_id=E.employee_id);
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/*19. Write a query in SQL to display job title and average salary of employees.*/
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SELECT job_title, AVG(salary)
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FROM employees
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NATURAL JOIN jobs
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GROUP BY job_title;
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@ -0,0 +1,8 @@
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/*2. Write a query in SQL to display the first and last name, department, city, and state province for each employee. */
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SELECT E.first_name,E.last_name,
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D.department_name, L.city, L.state_province
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FROM employees E
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JOIN departments D
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ON E.department_id = D.department_id
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JOIN locations L
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ON D.location_id = L.location_id;
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/*20. Write a query in SQL to display the details of jobs which was done by any of the employees who is presently earning a
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salary on and above 12000.*/
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SELECT a.*
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FROM job_history a
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JOIN employees m
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ON (a.employee_id = m.employee_id)
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WHERE salary >= 12000;
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/*21. Write a query in SQL to display the country name, city, and number of those departments where at leaste 2 employees
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are working.*/
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SELECT country_name,city, COUNT(department_id)
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FROM countries
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JOIN locations USING (country_id)
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JOIN departments USING (location_id)
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WHERE department_id IN
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(SELECT department_id
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FROM employees
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GROUP BY department_id
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HAVING COUNT(department_id)>=2)
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GROUP BY country_name,city;
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/*22. Write a query in SQL to display the department name, full name (first and last name) of manager, and their city.*/
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SELECT department_name, first_name || ' ' || last_name AS name_of_manager, city
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FROM departments D
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JOIN employees E
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ON (D.manager_id=E.employee_id)
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JOIN locations L USING (location_id);
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@ -0,0 +1,5 @@
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/*23. Write a query in SQL to display the employee ID, job name, number of days worked in for all those jobs in department 80.*/
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SELECT employee_id, job_title, end_date-start_date DAYS
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FROM job_history
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NATURAL JOIN jobs
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WHERE department_id=80;
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@ -0,0 +1,7 @@
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/*24. Write a query in SQL to display the full name (first and last name), and salary of those employees who working in any
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department located in London.*/
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SELECT first_name || ' ' || last_name AS Employee_name, salary
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FROM employees
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JOIN departments USING (department_id)
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JOIN locations USING (location_id)
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WHERE city = 'London';
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@ -0,0 +1,9 @@
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/*25. Write a query in SQL to display full name(first and last name), job title, starting and ending date of last jobs for
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those employees with worked without a commission percentage.*/
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SELECT first_name || ' ' || last_name AS Employee_name,
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job_title, start_date, end_date
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FROM job_history a
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JOIN jobs b USING (job_id)
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JOIN employees c
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ON ( a.employee_id = c.employee_id)
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WHERE commission_pct IS NULL;
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@ -0,0 +1,5 @@
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/*26. Write a query in SQL to display the department name and number of employees in each of the department.*/
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SELECT department_name, COUNT(*)
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FROM employees
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NATURAL JOIN departments
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GROUP BY department_name;
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@ -0,0 +1,5 @@
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/*3. Write a query in SQL to display the first name, last name, salary, and job grade for all employees. */
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SELECT E.first_name, E.last_name, E.salary, J.grade_level
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FROM employees E
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JOIN job_grades J
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ON E.salary BETWEEN J.lowest_sal AND J.highest_sal;
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@ -0,0 +1,9 @@
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/*4. Write a query in SQL to display the first name, last name, department number and department name, for all employees for
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departments 80 or 40. */
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SELECT E.first_name , E.last_name ,
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E.department_id , D.department_name
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FROM employees E
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JOIN departments D
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ON E.department_id = D.department_id
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AND E.department_id IN (80 , 40)
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ORDER BY E.last_name;
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@ -0,0 +1,10 @@
|
||||
/*5. Write a query in SQL to display those employees who contain a letter z to their first name and also display their last name,
|
||||
department, city, and state province.*/
|
||||
SELECT E.first_name,E.last_name,
|
||||
D.department_name, L.city, L.state_province
|
||||
FROM employees E
|
||||
JOIN departments D
|
||||
ON E.department_id = D.department_id
|
||||
JOIN locations L
|
||||
ON D.location_id = L.location_id
|
||||
WHERE E.first_name LIKE '%z%';
|
||||
@ -0,0 +1,5 @@
|
||||
/*6. Write a query in SQL to display all departments including those where does not have any employee.*/
|
||||
SELECT E.first_name, E.last_name, E.department_id, D.department_name
|
||||
FROM employees E
|
||||
RIGHT OUTER JOIN departments D
|
||||
ON E.department_id = D.department_id;
|
||||
@ -0,0 +1,7 @@
|
||||
/*7. Write a query in SQL to display the first and last name and salary for those employees who earn less than the employee
|
||||
earn whose number is 182.*/
|
||||
SELECT E.first_name, E.last_name, E.salary
|
||||
FROM employees E
|
||||
JOIN employees S
|
||||
ON E.salary < S.salary
|
||||
AND S.employee_id = 182;
|
||||
@ -0,0 +1,6 @@
|
||||
/*8. Write a query in SQL to display the first name of all employees including the first name of their manager.*/
|
||||
SELECT E.first_name AS "Employee Name",
|
||||
M.first_name AS "Manager"
|
||||
FROM employees E
|
||||
JOIN employees M
|
||||
ON E.manager_id = M.employee_id;
|
||||
@ -0,0 +1,5 @@
|
||||
/*9. Write a query in SQL to display the department name, city, and state province for each department.*/
|
||||
SELECT D.department_name , L.city , L.state_province
|
||||
FROM departments D
|
||||
JOIN locations L
|
||||
ON D.location_id = L.location_id;
|
||||
@ -0,0 +1,6 @@
|
||||
/*2Write a SQL query to display the average price of items of each company, showing the name of the company.*/
|
||||
SELECT AVG(pro_price), company_mast.com_name
|
||||
FROM item_mast INNER JOIN company_mast
|
||||
ON item_mast.pro_com= company_mast.com_id
|
||||
GROUP BY company_mast.com_name
|
||||
HAVING AVG(pro_price) >= 350;
|
||||
@ -0,0 +1,10 @@
|
||||
/*3Write a SQL query to display the average price of items of each company, showing the name of the company.*/
|
||||
SELECT A.pro_name, A.pro_price, F.com_name
|
||||
FROM item_mast A INNER JOIN company_mast F
|
||||
ON A.pro_com = F.com_id
|
||||
AND A.pro_price =
|
||||
(
|
||||
SELECT MAX(A.pro_price)
|
||||
FROM item_mast A
|
||||
WHERE A.pro_com = F.com_id
|
||||
);
|
||||
@ -0,0 +1,5 @@
|
||||
/*Write a SQL query to display all the data from the item_mast, including all the data for each item's producer company.*/
|
||||
SELECT *
|
||||
FROM item_mast
|
||||
INNER JOIN company_mast
|
||||
ON item_mast.pro_com= company_mast.com_id;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a SQL query to display the average price of items of each company, showing the name of the company.*/
|
||||
SELECT AVG(pro_price), company_mast.com_name
|
||||
FROM item_mast INNER
|
||||
JOIN company_mast
|
||||
ON item_mast.pro_com= company_mast.com_id
|
||||
GROUP BY company_mast.com_name;
|
||||
@ -0,0 +1,5 @@
|
||||
/*Write a SQL query to display the item name, price, and company name of all the products.*/
|
||||
SELECT item_mast.pro_name, pro_price, company_mast.com_name
|
||||
FROM item_mast
|
||||
INNER JOIN company_mast
|
||||
ON item_mast.pro_com = company_mast.com_id;
|
||||
@ -0,0 +1,10 @@
|
||||
/*Write a SQL statement to find the details of a order i.e. order number, order date, amount of order, which customer gives the
|
||||
order and which salesman works for that customer and how much commission he gets for an order.*/
|
||||
SELECT a.ord_no,a.ord_date,a.purch_amt,
|
||||
b.cust_name AS "Customer Name", b.grade,
|
||||
c.name AS "Salesman", c.commission
|
||||
FROM orders a
|
||||
INNER JOIN customer b
|
||||
ON a.customer_id=b.customer_id
|
||||
INNER JOIN salesman c
|
||||
ON a.salesman_id=c.salesman_id;
|
||||
@ -0,0 +1,9 @@
|
||||
/*Write a SQL statement to find the list of customers who appointed a salesman for their jobs who does not live in the same
|
||||
city where their customer lives, and gets a commission is above 12%.*/
|
||||
SELECT a.cust_name AS "Customer Name",
|
||||
a.city, b.name AS "Salesman", b.city,b.commission
|
||||
FROM customer a
|
||||
INNER JOIN salesman b
|
||||
ON a.salesman_id=b.salesman_id
|
||||
WHERE b.commission>.12
|
||||
AND a.city<>b.city;
|
||||
@ -0,0 +1,8 @@
|
||||
/*Write a SQL statement to find the list of customers who appointed a salesman for their jobs who gets a commission from the
|
||||
company is more than 12%.*/
|
||||
SELECT a.cust_name AS "Customer Name",
|
||||
a.city, b.name AS "Salesman", b.commission
|
||||
FROM customer a
|
||||
INNER JOIN salesman b
|
||||
ON a.salesman_id=b.salesman_id
|
||||
WHERE b.commission>.12;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a SQL statement to know which salesman are working for which customer.*/
|
||||
SELECT a.cust_name AS "Customer Name",
|
||||
a.city, b.name AS "Salesman", b.commission
|
||||
FROM customer a
|
||||
INNER JOIN salesman b
|
||||
ON a.salesman_id=b.salesman_id;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a SQL statement to make a cartesian product between salesman and customer i.e. each salesman will appear for all
|
||||
customer and vice versa for that customer who belongs to a city. */
|
||||
SELECT *
|
||||
FROM salesman a
|
||||
CROSS JOIN customer b
|
||||
WHERE a.city IS NOT NULL;
|
||||
@ -0,0 +1,7 @@
|
||||
/*Write a SQL statement to make a cartesian product between salesman and customer i.e. each salesman will appear for all
|
||||
customer and vice versa for those salesmen who belongs to a city and the customers who must have a grade.*/
|
||||
SELECT *
|
||||
FROM salesman a
|
||||
CROSS JOIN customer b
|
||||
WHERE a.city IS NOT NULL
|
||||
AND b.grade IS NOT NULL;
|
||||
@ -0,0 +1,9 @@
|
||||
/*Write a SQL statement to make a cartesian product between salesman and customer i.e. each salesman will appear for all
|
||||
customer and vice versa for those salesmen who must belong a city which is not the same as his customer and the customers
|
||||
should have an own grade.*/
|
||||
SELECT *
|
||||
FROM salesman a
|
||||
CROSS JOIN customer b
|
||||
WHERE a.city IS NOT NULL
|
||||
AND b.grade IS NOT NULL
|
||||
AND a.city<>b.city;
|
||||
@ -0,0 +1,5 @@
|
||||
/*Write a SQL statement to make a cartesian product between salesman and customer i.e. each salesman will appear for all
|
||||
customer and vice versa.*/
|
||||
SELECT *
|
||||
FROM salesman a
|
||||
CROSS JOIN customer b;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a SQL statement to make a join on the tables salesman, customer and orders in such a form that the same column of each
|
||||
table will appear once and only the relational rows will come.*/
|
||||
SELECT *
|
||||
FROM orders
|
||||
NATURAL JOIN customer
|
||||
NATURAL JOIN salesman;
|
||||
@ -0,0 +1,13 @@
|
||||
/*Write a SQL statement to make a list for the salesmen who either work for one or more customers or yet to join any of the
|
||||
customer. The customer, may have placed, either one or more orders on or above order amount 2000 and must have a grade, or
|
||||
he may not have placed any order to the associated supplier.*/
|
||||
SELECT a.cust_name,a.city,a.grade,
|
||||
b.name AS "Salesman",
|
||||
c.ord_no, c.ord_date, c.purch_amt
|
||||
FROM customer a
|
||||
RIGHT OUTER JOIN salesman b
|
||||
ON b.salesman_id=a.salesman_id
|
||||
RIGHT OUTER JOIN orders c
|
||||
ON c.customer_id=a.customer_id
|
||||
WHERE c.purch_amt>=2000
|
||||
AND a.grade IS NOT NULL;
|
||||
@ -0,0 +1,10 @@
|
||||
/*Write a SQL statement to make a list for the salesmen who works either for one or more customer or not yet join under any of
|
||||
the customers who placed either one or more orders or no order to their supplier.*/
|
||||
SELECT a.cust_name,a.city,a.grade,
|
||||
b.name AS "Salesman",
|
||||
c.ord_no, c.ord_date, c.purch_amt
|
||||
FROM customer a
|
||||
RIGHT OUTER JOIN salesman b
|
||||
ON b.salesman_id=a.salesman_id
|
||||
RIGHT OUTER JOIN orders c
|
||||
ON c.customer_id=a.customer_id;
|
||||
@ -0,0 +1,9 @@
|
||||
/*Write a SQL statement to make a list in ascending order for the customer who holds a grade less than 300 and works either
|
||||
through a salesman or by own.*/
|
||||
SELECT a.cust_name,a.city,a.grade,
|
||||
b.name AS "Salesman", b.city
|
||||
FROM customer a
|
||||
LEFT OUTER JOIN salesman b
|
||||
ON a.salesman_id=b.salesman_id
|
||||
WHERE a.grade<300
|
||||
ORDER BY a.customer_id;
|
||||
@ -0,0 +1,7 @@
|
||||
/*Write a SQL statement to make a list in ascending order for the customer who works either through a salesman or by own.*/
|
||||
SELECT a.cust_name,a.city,a.grade,
|
||||
b.name AS "Salesman",b.city
|
||||
FROM customer a
|
||||
LEFT JOIN salesman b
|
||||
ON a.salesman_id=b.salesman_id
|
||||
order by a.customer_id;
|
||||
@ -0,0 +1,8 @@
|
||||
/*Write a SQL statement to make a list in ascending order for the salesmen who works either for one or more customer or not
|
||||
yet join under any of the customers.*/
|
||||
SELECT a.cust_name,a.city,a.grade,
|
||||
b.name AS "Salesman", b.city
|
||||
FROM customer a
|
||||
RIGHT OUTER JOIN salesman b
|
||||
ON b.salesman_id=a.salesman_id
|
||||
ORDER BY b.salesman_id;
|
||||
@ -0,0 +1,7 @@
|
||||
/*Write a SQL statement to make a list with order no, purchase amount, customer name and their cities for those orders which
|
||||
order amount between 500 and 2000.*/
|
||||
SELECT a.ord_no,a.purch_amt,
|
||||
b.cust_name,b.city
|
||||
FROM orders a,customer b
|
||||
WHERE a.customer_id=b.customer_id
|
||||
AND a.purch_amt BETWEEN 500 AND 2000;
|
||||
@ -0,0 +1,9 @@
|
||||
/*Write a SQL statement to make a report with customer name, city, order no. order date, purchase amount for only those customers
|
||||
on the list who must have a grade and placed one or more orders or which order(s) have been placed by the customer who is neither
|
||||
in the list not have a grade.*/
|
||||
SELECT a.cust_name,a.city, b.ord_no,
|
||||
b.ord_date,b.purch_amt AS "Order Amount"
|
||||
FROM customer a
|
||||
FULL OUTER JOIN orders b
|
||||
ON a.customer_id=b.customer_id
|
||||
WHERE a.grade IS NOT NULL;
|
||||
@ -0,0 +1,7 @@
|
||||
/*Write a SQL statement to make a report with customer name, city, order no. order date, purchase amount for those customers
|
||||
from the existing list who placed one or more orders or which order(s) have been placed by the customer who is not on the list*/
|
||||
SELECT a.cust_name,a.city, b.ord_no,
|
||||
b.ord_date,b.purch_amt AS "Order Amount"
|
||||
FROM customer a
|
||||
FULL OUTER JOIN orders b
|
||||
ON a.customer_id=b.customer_id;
|
||||
@ -0,0 +1,9 @@
|
||||
/* Write a SQL statement to make a report with customer name, city, order number, order date, and order amount in ascending
|
||||
order according to the order date to find that either any of the existing customers have placed no order or placed one or more
|
||||
orders.*/
|
||||
SELECT a.cust_name,a.city, b.ord_no,
|
||||
b.ord_date,b.purch_amt AS "Order Amount"
|
||||
FROM customer a
|
||||
LEFT OUTER JOIN orders b
|
||||
ON a.customer_id=b.customer_id
|
||||
order by b.ord_date;
|
||||
@ -0,0 +1,11 @@
|
||||
/*Write a SQL statement to make a report with customer name, city, order number, order date, order amount salesman name and
|
||||
commission to find that either any of the existing customers have placed no order or placed one or more orders by their salesman
|
||||
or by own. */
|
||||
SELECT a.cust_name,a.city, b.ord_no,
|
||||
b.ord_date,b.purch_amt AS "Order Amount",
|
||||
c.name,c.commission
|
||||
FROM customer a
|
||||
LEFT OUTER JOIN orders b
|
||||
ON a.customer_id=b.customer_id
|
||||
LEFT OUTER JOIN salesman c
|
||||
ON c.salesman_id=b.salesman_id;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a SQL statement to prepare a list with salesman name, customer name and their cities for the salesmen and customer who
|
||||
belongs to the same city.*/
|
||||
SELECT salesman.name AS "Salesman",
|
||||
customer.cust_name, customer.city
|
||||
FROM salesman,customer
|
||||
WHERE salesman.city=customer.city;
|
||||
@ -0,0 +1,4 @@
|
||||
/*Write a SQL statement that finds out each order number followed by the name of the customers who made the order. */
|
||||
SELECT orders.ord_no, customer.cust_name
|
||||
FROM orders, customer
|
||||
WHERE orders.customer_id = customer.customer_id;
|
||||
@ -0,0 +1,8 @@
|
||||
/*Write a SQL statement that produces all orders with the order number, customer name, commission rate and
|
||||
earned commission amount for those customers who carry their grade more than 200 and served by an existing salesman.*/
|
||||
SELECT ord_no, cust_name, commission AS "Commission%",
|
||||
purch_amt*commission AS "Commission"
|
||||
FROM salesman,orders,customer
|
||||
WHERE orders.customer_id = customer.customer_id
|
||||
AND orders.salesman_id = salesman.salesman_id
|
||||
AND customer.grade>=200;
|
||||
@ -0,0 +1,9 @@
|
||||
/*Write a SQL statement that shorts out the customer and their grade who made an order. Each of the customers
|
||||
must have a grade and served by at least a salesman, who belongs to a city.*/
|
||||
SELECT customer.cust_name AS "Customer",
|
||||
customer.grade AS "Grade"
|
||||
FROM orders, salesman, customer
|
||||
WHERE orders.customer_id = customer.customer_id
|
||||
AND orders.salesman_id = salesman.salesman_id
|
||||
AND salesman.city IS NOT NULL
|
||||
AND customer.grade IS NOT NULL;
|
||||
@ -0,0 +1,4 @@
|
||||
/*Write a SQL statement to find the names of all customers along with the salesmen who works for them.*/
|
||||
SELECT customer.cust_name, salesman.name
|
||||
FROM customer,salesman
|
||||
WHERE salesman.salesman_id = customer.salesman_id;
|
||||
@ -0,0 +1,6 @@
|
||||
/*Write a query to find those customers with their name and those salesmen with their name and city
|
||||
who lives in the same city.*/
|
||||
SELECT customer.cust_name,
|
||||
salesman.name, salesman.city
|
||||
FROM salesman, customer
|
||||
WHERE salesman.city = customer.city;
|
||||
Some files were not shown because too many files have changed in this diff Show More
Loading…
Reference in New Issue
Block a user