MySQL - 11GROUP BY与HAVING的使用
group by的应用场景
我们已经掌握使用select语句结合where查询条件获取需要的数据，但在实际的应用中，还会遇到 下面这类需求，又该如何解决?
- 公司想知道每个部门有多少名员工
- 班主任想统计各科第 一名的成绩
- 某门店想掌握男、女性会员的人数及平均年龄
group by的使用
从字面上理解，group by表示根据某种规则对数据进行分组，它必须配合聚合函数进行使用，对数据进行分组后可以进行count、sum、avg、max和min等运算。
#group by语法
SELECT column_name, aggregate_function(column_name) FROM table_name
GROUP BY column_name
说明:
- aggregate_function表示聚合函数。
- group by可以对一列或多列进行分组。
having的使用
在 SQL 中增加 HAVING 子句原因是，WHERE 关键字无法与聚合函数一起使用。HAVING 子句可 以对分组后的各组数据进行筛选。
# having语法
SELECT column_name, aggregate_function(column_name) FROM table_name
WHERE column_name operator value
GROUP BY column_name
HAVING aggregate_function(column_name) operator value
实战案例
初始化表数据
create table employee(
id int not null auto_increment primary key,
name varchar(30),
sex varchar(1),
salary int,
dept varchar(30)
);
insert into employee(name, sex, salary, dept) values('lily0', '1', 5500, 'deptA');
insert into employee(name, sex, salary, dept) values('lily1', '0', 4500, 'deptA');
insert into employee(name, sex, salary, dept) values('lily2', '0', 4200, 'deptB');
insert into employee(name, sex, salary, dept) values('lily3', '1', 7500, 'deptB');
insert into employee(name, sex, salary, dept) values('lily4', '0', 8500, 'deptA');
insert into employee(name, sex, salary, dept) values('lily5', '1', 6800, 'deptB');
insert into employee(name, sex, salary, dept) values('lily6', '1', 12000, 'deptA');
insert into employee(name, sex, salary, dept) values('lily7', '1', 3500, 'deptA');
insert into employee(name, sex, salary, dept) values('lily8', '1', 6000, 'deptC');
insert into employee(name, sex, salary, dept) values('lily9', '1', 8000, 'deptC');
insert into employee(name, sex, salary, dept) values('lily10', '0', 10000, 'deptA');
insert into employee(name, sex, salary, dept) values('lily11', '0', 4000, 'deptA');
group by的使用
mysql> select * from employee;
+----+--------+------+--------+-------+
| id | name | sex | salary | dept |
+----+--------+------+--------+-------+
| 1 | lily0 | 1 | 5500 | deptA |
| 2 | lily1 | 0 | 4500 | deptA |
| 3 | lily2 | 0 | 4200 | deptB |
| 4 | lily3 | 1 | 7500 | deptB |
| 5 | lily4 | 0 | 8500 | deptA |
| 6 | lily5 | 1 | 6800 | deptB |
| 7 | lily6 | 1 | 12000 | deptA |
| 8 | lily7 | 1 | 3500 | deptA |
| 9 | lily8 | 1 | 6000 | deptC |
| 10 | lily9 | 1 | 8000 | deptC |
| 11 | lily10 | 0 | 10000 | deptA |
| 12 | lily11 | 0 | 4000 | deptA |
+----+--------+------+--------+-------+
12 rows in set (0.00 sec)
mysql> select sex, count(*) from employee group by sex;
+------+----------+
| sex | count(*) |
+------+----------+
| 1 | 7 |
| 0 | 5 |
+------+----------+
2 rows in set (0.00 sec)
mysql> select dept, count(*) from employee group by dept;
+-------+----------+
| dept | count(*) |
+-------+----------+
| deptA | 7 |
| deptB | 3 |
| deptC | 2 |
+-------+----------+
3 rows in set (0.01 sec)
mysql>
mysql> select dept, sum(salary) from employee group by dept;
+-------+-------------+
| dept | sum(salary) |
+-------+-------------+
| deptA | 48000 |
| deptB | 18500 |
| deptC | 14000 |
+-------+-------------+
3 rows in set (0.01 sec)
mysql> select dept, max(salary) from employee group by dept;
+-------+-------------+
| dept | max(salary) |
+-------+-------------+
| deptA | 12000 |
| deptB | 7500 |
| deptC | 8000 |
+-------+-------------+
3 rows in set (0.00 sec)
mysql> select dept, min(salary) from employee group by dept;
+-------+-------------+
| dept | min(salary) |
+-------+-------------+
| deptA | 3500 |
| deptB | 4200 |
| deptC | 6000 |
+-------+-------------+
3 rows in set (0.00 sec)
having的使用
mysql> select dept, count(*) from employee group by dept having count(*) < 3;
+-------+----------+
| dept | count(*) |
+-------+----------+
| deptC | 2 |
+-------+----------+
1 row in set (0.00 sec)
mysql> select dept, max(salary) from employee group by dept having max(salary) > 10000;
+-------+-------------+
| dept | max(salary) |
+-------+-------------+
| deptA | 12000 |
+-------+-------------+
1 row in set (0.00 sec)