最新SQL 面试用题
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SQL 面试用题是面试中最常见的题型 ,围绕五张表深入了解SQL面试题型,提升SQL编写水平
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
