使用不同的where条件计算相同的列

Question

我有一个名为employees的表，其架构如下所示：

mysql> desc employees;
+------------+---------------+------+-----+---------+-------+
| Field      | Type          | Null | Key | Default | Extra |
+------------+---------------+------+-----+---------+-------+
| emp_no     | int(11)       | NO   | PRI | NULL    |       |
| birth_date | date          | NO   |     | NULL    |       |
| first_name | varchar(14)   | NO   |     | NULL    |       |
| last_name  | varchar(16)   | NO   |     | NULL    |       |
| gender     | enum('M','F') | NO   |     | NULL    |       |
| hire_date  | date          | NO   |     | NULL    |       |
+------------+---------------+------+-----+---------+-------+

我只想查询男性（emp_no和女性（where generder = 'M')）的where generder = 'F'计数。

我想要的结果如下：

+--------+--------+
| Male   | FeMale |
+--------+--------+
| 300    | 120    |
+--------+--------+

有关如何在单个查询中执行此操作的任何建议吗？

Answer 1

SELECT SUM(gender = 'M') AS Male, SUM(gender = 'F') AS Female
FROM employees

Answer 2

select sum(case when gender = 'm' then 1 else 0 end) male,
sum(case when gender = 'f' then 1 else 0 end) female,
from employees

你可以尝试类似的东西