Mysql8.0习题系列(五):子查询(一篇搞懂子查询做题技巧,超详细)
Mysql8.0习题系列
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文章目录
- Mysql8.0习题系列
- 子查询
- 1.查询和Zlotkey相同部门的员工姓名和工资
- 2.查询工资比公司平均工资高的员工的员工号,姓名和工资。
- 3.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
- 4.查询和姓名中包含字母u的员工在相同部门的员工的员工号和姓名
- 5.查询在部门的location_id为1700的部门工作的员工的员工号
- 6.查询管理者是King的员工姓名和工资
- 7.查询工资最低的员工信息: last_name, salary
- 8.查询平均工资最低的部门信息
- 9.查询平均工资最低的部门信息和该部门的平均工资(相关子查询)
- 10.查询平均工资最高的 job 信息
- 11.查询平均工资高于公司平均工资的部门有哪些?
- 12.查询出公司中所有 manager 的详细信息
- 13.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?
- 14.查询平均工资最高的部门的 manager 的详细信息: last_name, department_id, email, salary
- 15. 查询部门的部门号,其中不包括job_id是"ST_CLERK"的部门号
- 16. 选择所有没有管理者的员工的last_name
- 17.查询员工号、姓名、雇用时间、工资,其中员工的管理者为 'De Haan'
- 18.查询各部门中工资比本部门平均工资高的员工的员工号, 姓名和工资(相关子查询)
- 19.查询每个部门下的部门人数大于 5 的部门名称(相关子查询)
- 20.查询每个国家下的部门个数大于 2 的国家编号(相关子查询)
子查询
1.查询和Zlotkey相同部门的员工姓名和工资
首先我们查出Zlotkey所在的部门,再放入子查询中进行选择
SELECT last_name,salary
FROM employees
WHERE department_id = (SELECT department_id FROM employees
WHERE last_name = 'Zlotkey'
);
2.查询工资比公司平均工资高的员工的员工号,姓名和工资。
先找出平均工资
SELECT employee_id,last_name, salary
FROM employees
WHERE salary > (SELECT AVG(salary) FROM employees);
3.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 > (SELECT MAX(salary) FROM employees WHERE job_id = 'SA_MAN');
4.查询和姓名中包含字母u的员工在相同部门的员工的员工号和姓名
SELECT employee_id, last_name
FROM employees
WHERE job_id IN (SELECT job_id FROM employees
WHERE last_name LIKE '%u%');
5.查询在部门的location_id为1700的部门工作的员工的员工号
SELECT
employee_id
FROM
employees
WHERE department_id IN
(SELECT
department_id
FROM
departments WHERE location_id = 1700);
6.查询管理者是King的员工姓名和工资
SELECT last_name, salary
FROM employees
WHERE manager_id IN
(SELECT employee_id FROM employees WHERE last_name = 'King'
);
7.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);
8.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
ORDER BY AVG(salary) LIMIT 0,1);
9.查询平均工资最低的部门信息和该部门的平均工资(相关子查询)
SELECT d.*, avg_sl.avg_dep FROM departments d,(SELECT department_id, AVG(salary) avg_dep FROM employees GROUP BY department_id
ORDER BY AVG(salary)) avg_sl
WHERE d.`department_id` = avg_sl.department_id
LIMIT 1
;
10.查询平均工资最高的 job 信息
SELECT * FROM jobs WHERE job_id = (SELECT job_id FROM employees GROUP BY job_id
ORDER BY AVG(salary) LIMIT 1);
11.查询平均工资高于公司平均工资的部门有哪些?
SELECT department_id FROM employees
WHERE department_id IS NOT NULL
GROUP BY department_id
HAVING AVG(salary) > (SELECT AVG(salary) FROM employees);
12.查询出公司中所有 manager 的详细信息
SELECT * FROM employees WHERE employee_id
IN (SELECT DISTINCT manager_id FROM employees);
13.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?
SELECT department_id, MAX(salary) FROM employees GROUP BY department_id
ORDER BY MAX(salary) ASC LIMIT 1;
14.查询平均工资最高的部门的 manager 的详细信息: last_name, department_id, email, salary
SELECT last_name, department_id, email,salary
FROM employees
WHERE manager_id = (SELECT DISTINCT employee_id FROM employees GROUP BY department_id ORDER BY AVG(salary) DESC
LIMIT 1);
15. 查询部门的部门号,其中不包括job_id是"ST_CLERK"的部门号
SELECT department_id
FROM departments d
WHERE department_id NOT IN (
SELECT DISTINCT department_id
FROM employees
WHERE job_id = 'ST_CLERK'
);
16. 选择所有没有管理者的员工的last_name
#方法1:
SELECT last_name
FROM employees
WHERE employee_id = (SELECT employee_id FROM employees WHERE manager_id IS NULL);
SELECT last_name
FROM employees e1
WHERE NOT EXISTS (
SELECT *
FROM employees e2
WHERE e1.manager_id = e2.employee_id
);
17.查询员工号、姓名、雇用时间、工资,其中员工的管理者为 ‘De Haan’
SELECT employee_id, last_name, hire_date, salary
FROM employees
WHERE manager_id = (
SELECT employee_id FROM employees WHERE last_name = 'De Haan'
);
18.查询各部门中工资比本部门平均工资高的员工的员工号, 姓名和工资(相关子查询)
SELECT employee_id, last_name, salary
FROM employees e, (SELECT department_id, AVG(salary) av_dep
FROM employees GROUP BY department_id) e2
WHERE e.`department_id` = e2.department_id
AND e.`salary` > av_dep;
19.查询每个部门下的部门人数大于 5 的部门名称(相关子查询)
方法一
SELECT department_name
FROM departments
WHERE department_id
IN (SELECT department_id
FROM employees GROUP BY department_id
HAVING COUNT(*) > 5);
方法二
SELECT department_name,department_id
FROM departments d
WHERE 5 < (
SELECT COUNT(*)
FROM employees e
WHERE d.`department_id` = e.`department_id`
);
20.查询每个国家下的部门个数大于 2 的国家编号(相关子查询)
SELECT country_id
FROM locations l
WHERE 2 <(
SELECT COUNT(*) FROM departments d
WHERE l.`location_id` = d.`location_id`);
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