Mysql习题系列(二):多表查询(一篇学会做Mysql多表查询题,超详细~)

Mysql8.0习题系列

软件下载地址
提取码:7v7u
数据下载地址
提取码:e6p9



1.多表查询1

1.1题目

1.显示所有员工的姓名,部门号和部门名称。
2.查询90号部门员工的job_id和90号部门的location_id
3.选择所有有奖金的员工的 last_name , department_name , location_id , city
4.选择city在Toronto工作的员工的 last_name , job_id , department_id , department_name
5.查询员工所在的部门名称、部门地址、姓名、工作、工资,其中员工所在部门的部门名称为’Executive’
6.选择指定员工的姓名,员工号,以及他的管理者的姓名和员工号,结果类似于下面的格式
employees Emp # manager Mgr#
kochhar 101 king 100 
7.查询哪些部门没有员工
8. 查询哪个城市没有部门
9. 查询部门名为 Sales 或 IT 的员工信息 

1.2答案

1.显示所有员工的姓名,部门号和部门名称。

SELECT 
  last_name,
  e.department_id,
  d.department_name 
FROM
  employees e 
  LEFT OUTER JOIN departments d 
    ON e.department_id = d.department_id ;

2.查询90号部门员工的job_id和90号部门的location_id

SELECT 
  job_id,
  location_id 
FROM
  employees e 
  INNER JOIN departments d 
    ON e.department_id = 90 ;

3.选择所有有奖金的员工的 last_name , department_name , location_id , city

SELECT last_name, department_name, d.location_id, city
FROM employees e
LEFT OUTER JOIN departments d
ON e.`department_id` = d.`department_id` 
LEFT OUTER JOIN locations l 
ON l.`location_id` = d.`location_id`
WHERE e.`commission_pct` IS NOT NULL;

4.选择city在Toronto工作的员工的 last_name , job_id , department_id , department_name

SELECT e.last_name, e.job_id,e.department_id,department_name
FROM employees e
JOIN departments d
ON e.`department_id` = d.`department_id`
JOIN locations l
ON d.`location_id` = l.`location_id`
WHERE city = 'Toronto';

5 查询员工所在的部门名称、部门地址、姓名、工作、工资,其中员工所在部门的部门名称为’Executive’

SELECT department_name, street_address, last_name, job_id, salary
FROM employees e
JOIN departments d
ON e.`department_id` = d.`department_id`
JOIN locations l
ON d.`location_id` = l.`location_id`
WHERE department_name ='Executive';

6.选择指定员工的姓名,员工号,以及他的管理者的姓名和员工号,结果类似于下面的格式

employees Emp # manager Mgr#
kochhar 101 king 100

SELECT e.last_name, e.employee_id 'emp#' , m.last_name manager , m.employee_id 'mgr#'
FROM employees e
JOIN employees m
ON e.`manager_id` = m.`employee_id` 
WHERE e.last_name = 'kochhar';

7.查询哪些部门没有员工。

思路:用外连接得到departments表与employees的连接关系,再筛选employee_id为null的departments表

SELECT d.department_id
FROM departments d
LEFT OUTER JOIN employees e
ON e.`department_id` = d.`department_id` 
WHERE employee_id IS NULL;

8.查询哪个城市没有部门 查询哪个城市没有部门,同样的思路,先查询location表,外连接departments表

SELECT city
FROM locations l
LEFT OUTER JOIN departments d
ON l.`location_id` = d.`location_id`
WHERE department_id IS NULL;

9. 查询部门名为 Sales 或 IT 的员工信息

查询部门名为sales或id的员工信息
SELECT * FROM 
employees e
JOIN departments d 
ON e.`department_id` = d.`department_id`
WHERE department_name IN ('Sales','IT');

2.多表查询2

2.1题目

#1.所有有门派的人员信息

#2.列出所有用户,并显示其机构信息

#3.列出所有门派

#4.所有不入门派的人员

#5.所有没人入的门派

#6.列出所有人员和机构的对照关系
MySQL Full Join的实现 。因为MySQL不支持FULL JOIN,下面是替代方法
#left join + union(可去除重复数据)+ right join

#7.列出所有没入派的人员和没人入的门派

首先创造数据集

CREATE DATABASE ex4;
CREATE TABLE `t_dept` (
`id` INT(11) NOT NULL AUTO_INCREMENT,
`deptName` VARCHAR(30) DEFAULT NULL,
`address` VARCHAR(40) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=INNODB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;
CREATE TABLE `t_emp` (
`id` INT(11) NOT NULL AUTO_INCREMENT,
`name` VARCHAR(20) DEFAULT NULL,
`age` INT(3) DEFAULT NULL,
`deptId` INT(11) DEFAULT NULL,
empno INT NOT NULL,
PRIMARY KEY (`id`),
KEY `idx_dept_id` (`deptId`)
#CONSTRAINT `fk_dept_id` FOREIGN KEY (`deptId`) REFERENCES `t_dept` (`id`)
) ENGINE=INNODB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;
INSERT INTO t_dept(deptName,address) VALUES('华山','华山');
INSERT INTO t_dept(deptName,address) VALUES('丐帮','洛阳');
INSERT INTO t_dept(deptName,address) VALUES('峨眉','峨眉山');
INSERT INTO t_dept(deptName,address) VALUES('武当','武当山');
INSERT INTO t_dept(deptName,address) VALUES('明教','光明顶');
INSERT INTO t_dept(deptName,address) VALUES('少林','少林寺');
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('风清扬',90,1,100001);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('岳不群',50,1,100002);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('令狐冲',24,1,100003);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('洪七公',70,2,100004);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('乔峰',35,2,100005);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('灭绝师太',70,3,100006);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('周芷若',20,3,100007);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('张三丰',100,4,100008);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('张无忌',25,5,100009);
INSERT INTO t_emp(NAME,age,deptId,empno) VALUES('韦小宝',18,NULL,100010);

2.2 答案

1.所有有门派的人员信息

SELECT * FROM
t_emp e
INNER  JOIN
t_dept d
ON e.`deptId` = d.id;

2.列出所有用户,并显示其机构信息

SELECT * FROM 
t_emp e LEFT JOIN t_dept d
ON e.`deptId` = d.`id`;

3.列出所有门派

SELECT * FROM t_dept;

4.所有不入门派的人员

SELECT * FROM t_emp e
LEFT OUTER JOIN t_dept d
ON e.`deptId` = d.`id`
WHERE deptId IS NULL;

5.所有没人入的门派

SELECT * FROM t_dept d
LEFT OUTER JOIN t_emp e
ON e.`deptId` = d.`id`
WHERE e.`deptId` IS NULL;

6.列出所有人员和机构的对照关系

MySQL Full Join的实现 。因为MySQL不支持FULL JOIN,下面是替代方法
#left join + union(可去除重复数据)+ right join

SELECT *
FROM t_emp e LEFT JOIN t_dept d
ON e.`deptId` = d.`id`
UNION
SELECT *
FROM t_emp e RIGHT JOIN t_dept d
ON e.deptId = d.id;

7.列出所有没入派的人员和没人入的门派

SELECT *
FROM t_emp e LEFT OUTER JOIN t_dept d
ON e.deptId = d.id
WHERE e.`deptId` IS NULL
UNION
SELECT *
FROM t_emp e RIGHT OUTER JOIN t_dept d
ON e.deptId = d.id
WHERE e.`deptId` IS NULL;
posted @ 2022-08-27 11:09  JOJO数据科学  阅读(177)  评论(0编辑  收藏  举报