在mysql中使用group by的平均值

Question

我有三张桌子

EMPLOYEE table with ID and NAME
COMPANY table with ID and NAME
and 
SALARY table with EMPLOYEE_ID, COMPANY_ID and SALARY

我想打印平均工资大于或等于 40000 的每个公司的名称

我的查询是

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;

Answer 1

您可以通过公司名称族且具有在条件having条款：

SELECT   c.name
FROM     company c
JOIN     salary s ON c.id = s.company_id
GROUP BY c.name
HAVING   AVG(salary) >= 40000

Answer 2

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

Answer 3

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;