MySQL多表查询讲解
2021-03-08
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1. 开始之前
-- ----------------------------
-- Table structure for dept
-- ----------------------------
DROP TABLE IF EXISTS `dept`;
CREATE TABLE `dept` (
`did` int(11) NOT NULL AUTO_INCREMENT COMMENT '部门ID',
`dname` varchar(60) DEFAULT NULL COMMENT '部门名称',
PRIMARY KEY (`did`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of dept
-- ----------------------------
BEGIN;
INSERT INTO `dept` VALUES (1, '研发部');
INSERT INTO `dept` VALUES (2, '人事部');
INSERT INTO `dept` VALUES (3, '测试部');
INSERT INTO `dept` VALUES (4, '销售部');
INSERT INTO `dept` VALUES (5, '生产部');
COMMIT;
-- ----------------------------
-- Table structure for employee
-- ----------------------------
DROP TABLE IF EXISTS `employee`;
CREATE TABLE `employee` (
`id` int(11) NOT NULL AUTO_INCREMENT COMMENT '主键',
`name` varchar(60) NOT NULL COMMENT '姓名',
`age` tinyint(4) DEFAULT NULL COMMENT '年龄',
`sex` tinyint(2) NOT NULL DEFAULT '1' COMMENT '性别，1男，2女',
`salary` decimal(10,2) NOT NULL COMMENT '薪资',
`hire_date` date NOT NULL COMMENT '聘用日期',
`dept_id` int(11) DEFAULT NULL COMMENT '部门ID',
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=12 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of employee
-- ----------------------------
BEGIN;
INSERT INTO `employee` VALUES (1, '菜虚鲲', 20, 2, 10000.00, '2020-01-10', 1);
INSERT INTO `employee` VALUES (2, '奥力给', 30, 1, 18000.00, '2020-01-08', 1);
INSERT INTO `employee` VALUES (3, '老八', 28, 1, 7000.00, '2020-01-07', 1);
INSERT INTO `employee` VALUES (4, '小张', 25, 1, 8000.00, '2020-01-10', 1);
INSERT INTO `employee` VALUES (5, '小红', 20, 2, 6000.00, '2020-01-05', 2);
INSERT INTO `employee` VALUES (6, '小丽', 23, 2, 6500.00, '2020-01-05', 2);
INSERT INTO `employee` VALUES (7, '小花', 21, 2, 5500.00, '2020-01-10', 2);
INSERT INTO `employee` VALUES (8, '马小跳', 25, 1, 7000.00, '2020-01-01', 3);
INSERT INTO `employee` VALUES (9, '张大骚', 30, 1, 9000.00, '2020-01-07', 3);
INSERT INTO `employee` VALUES (10, '马冬梅', 31, 2, 5000.00, '2020-01-07', 4);
INSERT INTO `employee` VALUES (11, '川坚果', 60, 1, 100.00, '2020-01-08', NULL);
COMMIT;
2. 多表联合查询
2.1 语法
select 字段1，字段2... from 表1，表2... [where 条件]
2.2 实战
注意:多表联合查询需要添加条件，否则会直接输出 左表*右表，这种结果称之为笛卡尔乘积。
集合A中的数据乘以集合B中的数据等于笛卡尔乘积
MySQL> select * from employee,dept;
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 2 | 人事部 |
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 3 | 测试部 |
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 4 | 销售部 |
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 5 | 生产部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 2 | 人事部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 3 | 测试部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 4 | 销售部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 5 | 生产部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 2 | 人事部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 3 | 测试部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 4 | 销售部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 5 | 生产部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 2 | 人事部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 3 | 测试部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 4 | 销售部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 5 | 生产部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 3 | 测试部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 4 | 销售部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 5 | 生产部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 1 | 研发部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 3 | 测试部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 4 | 销售部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 5 | 生产部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 1 | 研发部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 3 | 测试部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 4 | 销售部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 5 | 生产部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 1 | 研发部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 4 | 销售部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 5 | 生产部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 1 | 研发部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 2 | 人事部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 4 | 销售部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 5 | 生产部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 1 | 研发部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 2 | 人事部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 5 | 生产部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 1 | 研发部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 2 | 人事部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 3 | 测试部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 4 | 销售部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 5 | 生产部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
55 rows in set (0.00 sec)
正确的查询方式是:以两表中相互关联的字段作为查询条件进行查询。
mysql> select * from employee,dept where employee.dept_id = dept.did;
查询结果
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
10 rows in set (0.00 sec)
输出结果为两个集合的交集。
3. Innner内连接查询
3.1 语法
select 字段1，字段2... from 表1 inner join 表2 on [条件];
3.2 实战
查询员工和部门信息
select * from employee inner join dept on employee.dept_id = dept.did;
输出结果
mysql> select * from employee inner join dept on employee.dept_id = dept.did;
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
10 rows in set (0.04 sec)
其输出结果和多表联合查询一致。
如果附加其他条件，可以直接用and连接符连接在on语句的后面
mysql> select * from employee inner join dept on employee.dept_id = dept.did and employee.salary >= 10000;
输出结果
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
2 rows in set (0.00 sec)
4. Left左外连接查询
左外连接查询，即左表的数据全部显示。
4.1 语法
select * from 表1 left join 表2 on [条件];
4.2 实战
查询员工和部门的所有信息
select * from employee left join dept on employee.dept_id = dept.did;
输出结果
mysql> select * from employee left join dept on employee.dept_id = dept.did;
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | NULL | NULL |
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
11 rows in set (0.00 sec)
左表中的数据全部显示，右表中的数据只显示符合条件的，不符合条件的以NULL填充
更直观的展现上述关系
mysql> select * from employee left join dept on employee.dept_id = dept.did and dept.did = 1;
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | NULL | NULL |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | NULL | NULL |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | NULL | NULL |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | NULL | NULL |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | NULL | NULL |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | NULL | NULL |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | NULL | NULL |
+----+-----------+-----+-----+----------+------------+---------+------+-----------+
11 rows in set (0.01 sec)
5. Right右外连接查询
右外连接查询，与左外连接查询正好相反，即右表的数据全部显示。
5.1 语法
select * from 表1 right join 表2 on [条件];
5.2 实战
查询员工和部门信息
select * from employee right join dept on employee.dept_id = dept.did;
输出结果
mysql> select * from employee right join dept on employee.dept_id = dept.did;
+------+-----------+------+------+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+------+-----------+------+------+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | 5 | 生产部 |
+------+-----------+------+------+----------+------------+---------+-----+-----------+
11 rows in set (0.00 sec)
6. UNION全连接查询
全连接查询：其结果是在内连接查询的基础上显示左右两边没有的数据。
写法：左连接查询 UNION 右连接查询
6.1 语法
select * from 表1 left join 表2 on [条件]
union
select * from 表1 right join 表2 on [条件]
6.2 实战
以全连接的形式查询部门和员工数据
select * from employee left join dept on employee.dept_id = dept.did
union
select * from employee right join dept on employee.dept_id = dept.did;
输出结果
mysql> select * from employee left join dept on employee.dept_id = dept.did
-> union
-> select * from employee right join dept on employee.dept_id = dept.did;
+------+-----------+------+------+----------+------------+---------+------+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+------+-----------+------+------+----------+------------+---------+------+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 2 | 人事部 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 2 | 人事部 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 3 | 测试部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 4 | 销售部 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | NULL | NULL |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | 5 | 生产部 |
+------+-----------+------+------+----------+------------+---------+------+-----------+
12 rows in set (0.00 sec)
7. 嵌套查询
嵌套查询：在一个sql语句中使用多个select，第一次的查询结果可作为第二次查询结果的条件/表名使用。
7.1 作为表名使用
select * from (select id,name,age from employee) as em where em.id = 1;
查询结果
mysql> select * from (select id,name,age from employee) as em where em.id = 1;
+----+-----------+-----+
| id | name | age |
+----+-----------+-----+
| 1 | 菜虚鲲 | 20 |
+----+-----------+-----+
1 row in set (0.00 sec)
解释：把（select id,name,age from employee）语句的返回结果当作一个临时表，临时表的表名为em。
举例：查询平均工资在7000以上的部门
解题思路：
第一步，求部门平均工资
mysql> select AVG(salary) as ag,dept.dname from employee,dept where employee.dept_id = dept.did group by dept.did;
+--------------+-----------+
| ag | dname |
+--------------+-----------+
| 10750.000000 | 研发部 |
| 6000.000000 | 人事部 |
| 8000.000000 | 测试部 |
| 5000.000000 | 销售部 |
+--------------+-----------+
4 rows in set (0.00 sec)
第二步，把部门平均工资表当作临时表，进行查询。
select dname from
(
select
AVG(salary) as ag,
dept.dname
from employee,dept w
here employee.dept_id = dept.did
group by dept.did
) as dept_avg_salary
where dept_avg_salary.ag > 7000;
运行结果：
mysql> select dname from (select AVG(salary) as ag,dept.dname from employee,dept where employee.dept_id = dept.did group by dept.did) as dept_avg_salary where dept_avg_salary.ag > 7000;
+-----------+
| dname |
+-----------+
| 研发部 |
| 测试部 |
+-----------+
2 rows in set (0.00 sec)
7.2 作为查询条件使用
举例：找出工资最高的员工的所有信息
select * from employee,dept
where employee.dept_id = dept.did
and employee.salary = (select MAX(salary) from employee);
查询结果
mysql> select * from employee,dept where employee.dept_id = dept.did and employee.salary = (select MAX(salary) from employee);
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
1 row in set (0.01 sec)
举例：求工资大于所有人平均工资的员工的所有信息
select * from employee,dept
where employee.dept_id = dept.did
and employee.salary > (select AVG(salary) from employee);
查询结果
mysql> select * from employee,dept where employee.dept_id = dept.did and employee.salary > (select AVG(salary) from employee);
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| id | name | age | sex | salary | hire_date | dept_id | did | dname |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 1 | 研发部 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 1 | 研发部 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 3 | 测试部 |
+----+-----------+-----+-----+----------+------------+---------+-----+-----------+
4 rows in set (0.00 sec)
7.3. 嵌套查询关键字
7.3.1 ANY、SOME
两者意义相同，与比较运算符(=、>、>=、<、<=、<>)结合起来使用，any/some作用于子查询语句，只要子查询语句中有一个符合条件，就返回true。
select s1 from t1 where s1 > any (select s1 from t2);
假设子查询语句返回结果有三个result1，result2，result3，则上述语句等同于：
select s1 from t1 where s1 > result1 or s1 > result2 or s1 > result3;
7.3.2 IN
in 与 =any 相同。相当于:
select s1 from t1 where s1 = result1 or s1 = result2 or s1 = result3;
7.3.3 ALL
当子查询语句中的所有项都符合条件时，才返回true。
select s1 from t1 where s1 > all (select s1 from t2);
等同于:
select s1 from t1 where s1 > result1 and s1 > result2 and s1 > result3;
7.3.4 EXISTS、NOT EXISTS
语法:
select ... from tableName exists(subquery);
当子查询语句subquery返回列时，exists表达式为true，此时执行前面的查询语句。子查询语句没有返回任何列时，exists语句为false，不执行前面的查询语句。
mysql> select * from employee where exists (select * from employee where id =1);
+----+-----------+-----+-----+----------+------------+---------+
| id | name | age | sex | salary | hire_date | dept_id |
+----+-----------+-----+-----+----------+------------+---------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL |
+----+-----------+-----+-----+----------+------------+---------+
11 rows in set (0.00 sec)
当子查询语句没有返回任何列时
mysql> select * from employee where exists (select * from employee where id =12);
Empty set
select NULL 返回了列，所以:
mysql> select NULL;
+------+
| NULL |
+------+
| NULL |
+------+
1 row in set (0.00 sec)
mysql> select * from employee where exists (select NULL) and salary > 10000;
+----+-----------+-----+-----+----------+------------+---------+
| id | name | age | sex | salary | hire_date | dept_id |
+----+-----------+-----+-----+----------+------------+---------+
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 |
+----+-----------+-----+-----+----------+------------+---------+
1 row in set (0.00 sec)
ot exists与exists相反。
8 判断查询
8.1 IF
语法:
if(条件表达式,"结果为true","结果为false");
举例，将薪资大于大于8000的员工薪资级别设置为小康，小于8000设置为一般:
select *,if(salary > 8000,"小康","一般") as salary_level from employee;
输出结果:
mysql> select *,if(salary > 8000,"小康","一般") as salary_level from employee;
+----+-----------+-----+-----+----------+------------+---------+--------------+
| id | name | age | sex | salary | hire_date | dept_id | salary_level |
+----+-----------+-----+-----+----------+------------+---------+--------------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 小康 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 小康 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 一般 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 一般 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 一般 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 一般 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 一般 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 一般 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 小康 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 一般 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 一般 |
+----+-----------+-----+-----+----------+------------+---------+--------------+
11 rows in set (0.00 sec)
8.2 if … else if … else 形式
语法:
select ...,case when [条件1] then [result]
when [条件2] then [result]
else [result]
end
from tableName;
举例:
根据员工工资划分员工生活水平，小于7000为贫穷，7000 到 9000为一般，9000-10000为中等，10000-12000为中等偏上，大于12000为有钱。
mysql> select name,salary,case when salary < 7000 then '贫穷'
-> when salary < 9000 then '一般'
-> when salary < 10000 then '中等'
-> when salary < 12000 then '中等偏上'
-> else '有钱'
-> end as living_standard
-> from employee;
+-----------+----------+-----------------+
| name | salary | living_standard |
+-----------+----------+-----------------+
| 菜虚鲲 | 10000.00 | 中等偏上 |
| 奥力给 | 18000.00 | 有钱 |
| 老八 | 7000.00 | 一般 |
| 小张 | 8000.00 | 一般 |
| 小红 | 6000.00 | 贫穷 |
| 小丽 | 6500.00 | 贫穷 |
| 小花 | 5500.00 | 贫穷 |
| 马小跳 | 7000.00 | 一般 |
| 张大骚 | 9000.00 | 中等 |
| 马冬梅 | 5000.00 | 贫穷 |
| 川坚果 | 100.00 | 贫穷 |
+-----------+----------+-----------------+
11 rows in set (0.00 sec)
其形式类似于if…else if…else if…else形式。
8.3 switch case形式
语法:
select ..., case s1
when [value1] then [result1]
when [value2] then [result2]
when [value3] then [result3]
else [resultOther] end
from tableName;
举例，直接显示出员工性别:
mysql> select *,case sex
-> when 1 then '男'
-> when 2 then '女'
-> else '未知' end as employee_sex
-> from employee;
+----+-----------+-----+-----+----------+------------+---------+--------------+
| id | name | age | sex | salary | hire_date | dept_id | employee_sex |
+----+-----------+-----+-----+----------+------------+---------+--------------+
| 1 | 菜虚鲲 | 20 | 2 | 10000.00 | 2020-01-10 | 1 | 女 |
| 2 | 奥力给 | 30 | 1 | 18000.00 | 2020-01-08 | 1 | 男 |
| 3 | 老八 | 28 | 1 | 7000.00 | 2020-01-07 | 1 | 男 |
| 4 | 小张 | 25 | 1 | 8000.00 | 2020-01-10 | 1 | 男 |
| 5 | 小红 | 20 | 2 | 6000.00 | 2020-01-05 | 2 | 女 |
| 6 | 小丽 | 23 | 2 | 6500.00 | 2020-01-05 | 2 | 女 |
| 7 | 小花 | 21 | 2 | 5500.00 | 2020-01-10 | 2 | 女 |
| 8 | 马小跳 | 25 | 1 | 7000.00 | 2020-01-01 | 3 | 男 |
| 9 | 张大骚 | 30 | 1 | 9000.00 | 2020-01-07 | 3 | 男 |
| 10 | 马冬梅 | 31 | 2 | 5000.00 | 2020-01-07 | 4 | 女 |
| 11 | 川坚果 | 60 | 1 | 100.00 | 2020-01-08 | NULL | 男 |
+----+-----------+-----+-----+----------+------------+---------+--------------+
11 rows in set (0.00 sec)
