数据库 SQL语句

Question

雇员数据库如下：employee(employee_name,street,city) works(employee_name,company_name,salary) company(company_name,city)manages(employee_name,manager_name) 为查询语句写出SQL表达式：1,找出所居住街道城市与其经理相同的雇员2，找出工资总和最小的公司3，找出工资高于所在公司平均工资的雇员

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Answer 1

参考下：

1.select employee_name from employee where employee_name not in(select manager_name

from manages) and street in (select street from employee join manages on

employee.employee_name=manages.employee_name)
and city in (select city from employee join manages on

employee.employee_name=manages.employee_name)

2.select top 1 company_Name,sum(salary) from work group by company_name order by

sum(salary)

3.select employee_name from work join
(select Company_name,avg(salary) as avg_salary from work group by company_name)

tb on work.company_name=tb.company_name
where work.salary>avg_salary

Answer 2

1.select e.* form employee e,employee e1,manages m
where e1.employee_name=m.manager_name and e1.city=e.city and e.employee_name=m.employee_name
2.select *
from company
where company_name=
(select company_name from works where min(sum(salary))
3.select *
from employee
where salary>avg(salary)

Answer 3

1,
SELECT *
FROM EMPLOYEE
WHERE STREET,CITY =
(SELECT STREET,CITY
FROM EMPLOYEE,MANAGES
WHERE MANAGES.EMPLOYEE_NAME = EMPLOYEE.NAME)
2,(问得就有问题)
SELECT COMPANY_NAME,SUM(SALARY)
FROM WORKS
GROUP BY COMPANY_NAME
3，
建议自己写存储过程解决本回答被网友采纳

Answer 4

1.select e.* form employee e,employee e1,manages m
where e1.employee_nanme=m.manager_name and e1.city=e.city and e.employee_name=m.employee_name

2.select company_name,sum(salary)
from works
group by company_name