I have three table
EMPLOYEE table with ID and NAME
COMPANY table with ID and NAME
and
SALARY table with EMPLOYEE_ID, COMPANY_ID and SALARY
I wanted to print the name of every company where the average salary is greater than or equal to 40000
My query is
Select distinct COMPANY.NAME from COMPANY,SALARY,EMPLOYEE WHERE SALARY.COMPANY_ID=COMPANY.ID and (select avg(SALARY) from SALARY,EMPLOYEE WHERE SALARY.EMPLOYEE_ID=EMPLOYEE.ID) >=40000;
You could group by the company name and have the condition in the having clause:
SELECT c.name
FROM company c
JOIN salary s ON c.id = s.company_id
GROUP BY c.name
HAVING AVG(salary) >= 40000
SELECT DISTINCT COMPANY.NAME
FROM COMPANY
INNER JOIN SALARY ON COMPANY.ID = SALARY.COMPANY_ID
INNER JOIN EMPLOYEE ON EMPLOYEE.ID = SALARY.EMPLOYEE_ID
GROUP BY COMPANY.NAME HAVING AVG(SAL)>=40000
SELECT T1.NAME FROM COMPANY AS T1 INNER JOIN (SELECT COMPANY_ID, AVG(SALARY) AS avgSalary FROM SALARY GROUP BY COMPANY_ID) AS T2 ON T1.ID = T2.COMPANY_ID WHERE T2.avgSalary >= 40000;
