快速提升sql能力

1. 查询每个月倒数第 2 天入职的员工的信息.

2. 查询出 last_name 为 ‘Chen’ 的 manager 的信息.

3. 查询平均工资高于 8000 的部门 id 和它的平均工资.

4. 查询工资最低的员工信息: last_name, salary

5. 查询平均工资最低的部门信息

6. 查询平均工资最低的部门信息和该部门的平均工资

7. 查询平均工资最高的 job 信息

8. 查询平均工资高于公司平均工资的部门有哪些?

9. 查询出公司中所有 manager 的详细信息.

10. 各个部门中 最高工资中最低的那个部门的 最低工资是多少

11. 查询平均工资最高的部门的 manager 的详细信息: last_name,
    department_id, email, salary

12. 查询 1999 年来公司的人所有员工的最高工资的那个员工的信息.

13. 返回其它部门中比 job_id 为‘IT_PROG’部门所有工资都低的员工的
    员工号、姓名、job_id 以及 salary

***对应答案

1. 查询每个月倒数第 2 天入职的员工的信息.

select last_name, hire_date from employees where hire_date = last_day(hire_date) – 1

2. 查询出 last_name 为 ‘Chen’ 的 manager 的信息.
   1). 通过两条 sql 查询:

   select manager_id from employees
   where lower(last_name) = ‘chen’ --返回的结果为 108

   select *    from employees    where employee_id = 108     
   

2). 通过一条 sql 查询(自连接):

select m.*    from employees e, employees m    where e.manager_id = m.employee_id and e.last_name = 'Chen'       

3). 通过一条 sql 查询(子查询):

select * from employees where employee_id = ( select manager_id from employees where last_name = ‘Chen’ )

3. 查询平均工资高于 8000 的部门 id 和它的平均工资.

SELECT department_id, avg(salary) FROM employees e GROUP BY department_id HAVING avg(salary) > 8000

4. 查询工资最低的员工信息: last_name, salary

SELECT last_name, salary FROM employees WHERE salary = ( SELECT min(salary) FROM employees )

5. 查询平均工资最低的部门信息

SELECT * FROM departments WHERE department_id = ( SELECT department_id FROM

employees GROUP BY department_id HAVING avg(salary) = ( SELECT min(avg(salary))

FROM employees   GROUP BY department_id  )  ) 

6. 查询平均工资最低的部门信息和该部门的平均工资

select d.*, (select avg(salary) from employees where department_id = d.department_id)

from departments d where d.department_id = ( SELECT department_id FROM employees GROUP

BY department_id HAVING avg(salary) = ( SELECT min(avg(salary)) FROM employees

 GROUP BY department_id      )        )    

7. 查询平均工资最高的 job 信息

1). 按 job_id 分组, 查询最高的平均工资

SELECT max(avg(salary)) FROM employees GROUP BY job_id

2). 查询出平均工资等于 1) 的 job_id

SELECT job_id FROM employees GROUP BY job_id HAVING avg(salary) = ( SELECT

max(avg(salary)) FROM employees GROUP BY job_id )

3). 查询出 2) 对应的 job 信息

SELECT * FROM jobs
WHERE job_id = ( SELECT job_id FROM employees GROUP BY job_id HAVING avg(salary)

= ( SELECT max(avg(salary)) FROM employees GROUP BY job_id ) )

8. 查询平均工资高于公司平均工资的部门有哪些?

1). 查询出公司的平均工资

SELECT avg(salary) FROM employees

2). 查询平均工资高于 1) 的部门 ID

SELECT department_id FROM employees GROUP BY department_id HAVING avg(salary) > (

SELECT avg(salary) FROM employees )

9. 查询出公司中所有 manager 的详细信息.
   1). 查询出所有的 manager_id

SELECT distinct manager_id FROM employeess

2). 查询出 employee_id 为 1) 查询结果的那些员工的信息

SELECT employee_id, last_name FROM employees

WHERE employee_id in ( SELECT distinct manager_id FROM employees )

10. 各个部门中 最高工资中最低的那个部门的 最低工资是多少
    1). 查询出各个部门的最高工资

SELECT max(salary) FROM employees GROUP BY department_id

2). 查询出 1) 对应的查询结果的最低值: 各个部门中最低的最高工
资(无法查询对应的 department_id)

SELECT min(max(salary)) FROM employees GROUP BY department_id

3). 查询出 2) 所对应的部门 id 是多少: 各个部门中最高工资等于
2) 的那个部门的 id

SELECT department_id FROM employees GROUP BY department_id

HAVING max(salary) = ( SELECT min(max(salary)) FROM employees

GROUP BY department_id )

4). 查询出 3) 所在部门的最低工资

SELECT min(salary) FROM employees WHERE department_id = ( SELECT department_id

FROM employees GROUP BY department_id HAVING max(salary) = (

SELECT min(max(salary)) FROM employees GROUP BY department_id ) )

11. 查询平均工资最高的部门的 manager 的详细信息: last_name,

department_id, email, salary

1). 各个部门中, 查询平均工资最高的平均工资是多少

SELECT max(avg(salary)) FROM employees GROUP BY department_id

2). 各个部门中, 平均工资等于 1) 的那个部门的部门号是多少

SELECT department_id FROM employees GROUP BY department_id

HAVING avg(salary) = ( SELECT max(avg(salary)) FROM employees

GROUP BY department_id )

3). 查询出 2) 对应的部门的 manager_id

SELECT manager_id FROM departments WHERE department_id = ( SELECT department_id

FROM employees GROUP BY department_id HAVING avg(salary) = ( SELECT

max(avg(salary)) FROM employees GROUP BY department_id ) )

4). 查询出 employee_id 为 3) 查询的 manager_id 的员工的

last_name, department_id, email, salary SELECT last_name, department_id, email, salary

FROM employees WHERE employee_id = ( SELECT manager_id FROM departments

WHERE department_id = ( SELECT department_id FROM employees GROUP BY

department_id HAVING avg(salary) = ( SELECT max(avg(salary)) FROM employees

GROUP BY department_id ) ) )

12. 查询 1999 年来公司的人所有员工的最高工资的那个员工的信息.
    1). 查询出 1999 年来公司的所有的员工的 salary

SELECT salary FROM employees WHERE to_char(hire_date, ‘yyyy’) = ‘1999’

2). 查询出 1) 对应的结果的最大值

SELECT max(salary) FROM employees WHERE to_char(hire_date, ‘yyyy’) = ‘1999’

3). 查询工资等于 2) 对应的结果且 1999 年入职的员工信息

SELECT * FROM employees WHERE to_char(hire_date, ‘yyyy’) = ‘1999’ AND salary = (

SELECT max(salary) FROM employees WHERE to_char(hire_date, ‘yyyy’) = ‘1999’ )

13. 返回其它部门中比 job_id 为‘IT_PROG’部门所有工资都低的员工的员
    工号、姓名、job_id 以及 salary

SELECT employee_id, last_name, job_id, salary FROM employees WHERE salary < ALL

  (SELECT salary   FROM   employees   WHERE  job_id = 'IT_PROG') AND    job_id <> 

‘IT_PROG’;

高级子查询

• 书写多列子查询
• 在 FROM 子句中使用子查询
• 在 SQL 中使用单列子查询
• 书写相关子查询
• 使用 EXISTS 和 NOT EXISTS 操作符
• 使用子查询更新和删除数据
• 使用 WITH 子句
–多列子查询（不成对比较 & 成对比较） 1. 查询与 141 号或 174 号员工的 manager_id 和 department_id 相同的
其他员工的 employee_id, manager_id, department_id
[方式一]

SELECT employee_id, manager_id, department_id FROM employees WHERE manager_id IN

(SELECT manager_id FROM employees WHERE employee_id IN (174,141)) AND

department_id IN (SELECT department_id FROM employees WHERE employee_id IN

(174,141)) AND employee_id NOT IN(174,141);

[方式二]

SELECT employee_id, manager_id, department_id FROM employees WHERE (manager_id,

department_id) IN SELECT manager_id, department_id FROM employees

WHERE employee_id IN (141,174)) AND employee_id NOT IN (141,174);

–在 FROM 子句中使用子查询
2. 返回比本部门平均工资高的员工的 last_name, department_id,
salary 及平均工资

[方式一] select last_name,department_id,salary, (select avg(salary)from employees e3

where e1.department_id = e3.department_id group by department_id) avg_salary from

employees e1 where salary > (select avg(salary) from employees e2

       where e1.department_id = e2.department_id    --group by department_id           ) 

[方式二] SELECT a.last_name, a.salary, a.department_id, b.salavg

FROM employees a, (SELECT department_id, AVG(salary) salavg

FROM employees GROUP BY department_id) b WHERE a.department_id = b.department_id

AND a.salary > b.salavg;

–单列子查询表达式
• Oracle8i 只在下列情况下可以使用, 例如:
– SELECT 语句 (FROM 和 WHERE 子句)
– INSERT 语句中的 VALUES 列表中
• Oracle9i 中单列子查询表达式可在下列情况下使用:
– DECODE 和 CASE
– SELECT 中除 GROUP BY 子句以外的所有子句中
3. 显式员工的 employee_id,last_name 和 location。其中，若员工
department_id 与 location_id 为 1800 的 department_id 相同，则
location 为’Canada’,其余则为’USA’。

SELECT employee_id, last_name, (CASE department_id WHEN (SELECT department_id FROM

departments WHERE location_id = 1800) THEN ‘Canada’ ELSE ‘USA’ END) location FROM

employees;

4. 查询员工的 employee_id,last_name, 要求按照员工的
   department_name 排序

SELECT employee_id, last_name FROM employees e

ORDER BY (SELECT department_name FROM departments d WHERE e.department_id =

d.department_id);

–相关子查询

5.查询员工中工资大于本部门平均工资的员工的 last_name,
salary 和其 department_id

SELECT last_name, salary, department_id FROM employees outer WHERE salary >

(SELECT AVG(salary) FROM employees WHERE department_id =

  outer.department_id) ; 

6. 若 employees表中 employee_id与 job_history表中 employee_id
   相同的数目不小于 2，输出这些相同 id 的员工的 employee_id,last_name
   和其 job_id

SELECT e.employee_id, last_name,e.job_id FROM employees e WHERE 2 <= (SELECT COUNT(*)

FROM job_history WHERE employee_id = e.employee_id);

–EXISTS 操作符
• EXISTS 操作符检查在子查询中是否存在满足条件的行
• 如果在子查询中存在满足条件的行:
– 不在子查询中继续查找
– 条件返回 TRUE
7. 查询公司管理者的 employee_id,last_name,job_id,
department_id 信息

SELECT employee_id, last_name, job_id, department_id FROM employees outer WHERE EXISTS

( SELECT ‘X’ FROM employees WHERE manager_id = outer.employee_id);

8. 查询 departments 表 中， 不存 在于 employees 表 中的 部门 的

department_id 和 department_name

SELECT department_id, department_name FROM departments d WHERE NOT EXISTS

(SELECT ‘X’ FROM employees WHERE department_id = d.department_id);

–关于数据更新
9.修改表 employees,添加 department_name 列，赋予 department_id
相应的部门名称。

ALTER TABLE employees ADD(department_name VARCHAR2(14));

UPDATE employees e SET department_name = (SELECT department_name FROM

departments d WHERE e.department_id = d.department_id);

–关于数据删除
10.删除表 employees 中，其与 emp_history 表皆有的数据

DELETE FROM employees E WHERE employee_id in (SELECT employee_id FROM

emp_history WHERE employee_id = E.employee_id);

–WITH 子句
11. 查询公司中各部门的总工资大于公司中各部门的平均总工资的部门信息 WITH

dept_costs AS ( SELECT d.department_name, SUM(e.salary) AS dept_total FROM

employees e, departments d WHERE e.department_id = d.department_id GROUP BY

d.department_name), avg_cost AS ( SELECT SUM(dept_total)/COUNT(*) AS dept_avg

FROM dept_costs) SELECT * FROM dept_costs WHERE dept_total >

(SELECT dept_avg FROM avg_cost) ORDER BY department_name;

附加题目：
12.查询员工的 last_name, department_id, salary.其中员工的
salary,department_id 与有奖金的任何一个员工的 salary,
department_id 相同即可

select last_name, department_id, salary from employees where

(salary,department_id) in ( select salary,department_id from employees where

commission_pct is not null )

13.选择工资大于所有 JOB_ID = 'SA_MAN’的员工的工资的员工的
last_name, job_id, salary

select last_name, job_id, salary from employees where salary > all(

select salary from employees where job_id = ‘SA_MAN’ )

14.选择所有没有管理者的员工的 last_name

select last_name from employees e1 where not exists (

select ‘A’ from employees e2 where e1.manager_id = e2.employee_id )
15. 查询 10，50，20 号部门的 job_id，department_id 并且
department_id 按 10，50，20 的顺序排列
Column dummy noprint;

select job_id , department_id ,1 dummy from employees where department_id = 10 union

select job_id , department_id , 2 from employees where department_id = 50 union

select job_id , department_id , 3 from employees where department_id= 20 order by 3