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);
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 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 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;
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 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
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;
ALTER TABLE `t_dept`
add CEO INT(11) ;
update t_dept set CEO=2 where id=1;
update t_dept set CEO=4 where id=2;
update t_dept set CEO=6 where id=3;
update t_dept set CEO=8 where id=4;
update t_dept set CEO=9 where id=5;
求各个门派对应的掌门人名称:
select * from t_dept as b left join t_emp as a on b.CEO=a.id;
求所有当上掌门人的平均年龄:
select avg(a.age) from t_emp a inner join t_dept b on a.id=b.CEO ;
求所有人物对应的掌门名称:
SELECT
c.`name`,
ab.`name` ceoname
FROM
t_emp c
LEFT JOIN ( SELECT b.id, a.NAME FROM t_emp a INNER JOIN t_dept b ON b.CEO = a.id ) ab ON c.deptId = ab.id