MySQL-多表查询练习

1. 首先创建练习所需要的数据表

储备:建表操作:
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;

2. 添加测试数据

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);

3. 题目以及答案

#1.所有有门派的人员信息 ( A、B两表共有)
select *
from t_emp a inner join t_dept b
on a.deptId = b.id;
#2.列出所有用户，并显示其机构信息 (A的全集)
select *
from t_emp a left outer join t_dept b
on a.deptId = b.id;
#3.列出所有门派 (B的全集)
select *
from  t_dept  b;
#4.所有不入门派的人员 (A的独有)
select *
from t_emp a left join t_dept b
on a.deptId = b.id
# 注意是is null，不是b.id = null
where b.id is null;
#5.所有没人入的门派 (B的独有)
select *
from t_dept b left join  t_emp a
on a.deptId = b.id
where a.deptId is null;
# 使用右连接也是可以的，只不过显示效果不符合常理
# select *
# from t_emp a right join t_dept b
# on a.deptId = b.id
# where a.id is null;

#6.列出所有人员和机构的对照关系(AB全有)
#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 a left outer join t_dept b
on a.deptId = b.id
union
select *
from t_emp a right join t_dept b
on a.deptId = b.id
where a.id is null;
#7.列出所有没入派的人员和没人入的门派 (A的独有+B的独有)
SELECT *
FROM t_emp A LEFT JOIN t_dept B
ON A.deptId = B.id
WHERE B.`id` IS NULL
UNION
SELECT *
FROM t_emp A RIGHT JOIN t_dept B
ON A.deptId = B.id
WHERE A.`deptId` IS NULL;

学习地址：https://www.bilibili.com/video/BV1iq4y1u7vj?p=29&spm_id_from=pageDriver&vd_source=564bf1e544930db6e3436599f8935e24