Question

我有这张桌子：

| name | gender | date             |
------------------------------------
| Foo  | male   | 2013-09-16 10:23 |
| Name | male   | 2013-09-16 09:10 |
| Red  | male   | 2013-09-15 09:10 |
| Bar  | female | 2013-09-15 10:10 |
etc...

我需要按gender过滤每周的每天访问次数。

所以我应该得到，如果算上男性：

1: 2 visits
2: 0 visits
3: 0 visits
4: 0 visits
5: 0 visits
6: 0 visits
7: 1 visit

查询应为：

SELECT FROM table
WHERE gender = 'male'
GROUP BY DAYOFWEEK

我的查询不起作用所以我在这里询问是否有人知道如何使其有效...

Answer 1

尝试

SELECT dayofweek(date) dayofweek, gender, count(*) count
FROM table
GROUP BY DAYOFWEEK(date), gender

编辑：

如果你想让星期一为1，星期日为7，你也可以

SELECT (dayofweek(date)+5)%7+1 dayofweek, gender, count(*) count
FROM table
GROUP BY (dayofweek(date)+5)%7+1, gender

如果你想避免if构造......

Answer 2

使用WEEKDAY()代替，它从星期一开始。

SELECT WEEKDAY(date) weekday, gender, count(*) count
FROM table
GROUP BY WEEKDAY(date), gender

如果您需要从索引1开始，请使用或WEEKDAY（）+ 1。

SELECT (WEEKDAY(date)+1) weekday, gender, count(*) count
FROM table
GROUP BY WEEKDAY(date), gender

计算每周的每一天的发生次数

2 个答案: