返回SQL LIKE＆GROUP BY结果

Question

我在数据库中有一个非常简单的表，其中包含3列名为people列。 例：

user           id             notes
john           01             has red hair, last logged in 02/04/12
tony           02             has brown hair, last logged in 04/03/12
brian          03             has brown hair, last logged in 03/06/13
amanda         04             has blonde hair, last logged in 05/07/14
…

如果我想按注释字段分组并进行计数，则第2行和第3行显示为1，因为登录日期不同; 我想做的是砍掉日期和合并并纯粹按头发颜色计算，例如，如果我运行查询：

SELECT `notes`, COUNT( `user` ) AS Count 
FROM `people` 
WHERE `notes` LIKE "%hair%" GROUP BY `notes`;

我得到的结果是：

+-----------------------------------------+-------+
|notes                                    | Count |
+-----------------------------------------+-------+
|has red hair, last logged in 02/04/12    |      1|
|has brown hair, last logged in 04/03/12  |      1|
|has brown hair, last logged in 03/06/13  |      1|
|has blonde hair, last logged in 05/07/14 |      1|
+-----------------------------------------+-------+

想要达到以下结果：

+-------------------+------+
|notes              |Count | 
+-------------------+------+
|has red hair       |     1|
|has brown hair     |     2|
|has blonde hair    |     1|
+-------------------+------+

这是可能吗？

干杯。

Answer 1

您可以使用SUBSTRING_INDEX来执行此操作，如下所示：

SELECT SUBSTRING_INDEX(`notes`, ',', 1), COUNT( `user` ) AS Count
FROM `people`
WHERE `notes` LIKE "%hair%"
GROUP BY SUBSTRING_INDEX(`notes`, ',', 1)

当然这是一个有点肮脏的解决方法。 更好的方法是分离头发颜色，但我意识到这可能并不总是可行的。

演示