我想这是一个反复出现的问题,但我没有找到合适的主题来指出正确的方向.
+---------+----------------+---------------------+-----------------+ | id(int) | author(string) | date(datetime) | message(string) | +---------+----------------+---------------------+-----------------+ | 1 | John | 2016-01-01 17:18:00 | I | | 2 | Mary | 2016-01-01 14:22:00 | Just | | 3 | John | 2016-01-01 09:02:00 | Want | | 4 | John | 2016-01-02 17:18:00 | To | | 5 | Mary | 2016-01-03 18:26:00 | Say | | 6 | John | 2016-01-03 10:42:00 | Hello | +---------+----------------+---------------------+-----------------+
我想得到什么:
+------------+------+------+ | day | Mary | John | +------------+------+------+ | 2016-01-01 | 1 | 2 | | 2016-01-02 | 0 | 1 | | 2016-01-03 | 1 | 1 | +------------+------+------+
到目前为止,我想到了:
SELECT DATE(date) as day,
(SELECT COUNT(id) FROM chat WHERE author = 'Mary') AS 'Mary'
(SELECT COUNT(id) FROM chat WHERE author = 'John') AS 'John'
FROM chat
GROUP BY day
ORDER BY day ASC
但这给了我每一行每一作者的总消息数:
+------------+------+------+ | day | Mary | John | +------------+------+------+ | 2016-01-01 | 2 | 4 | | 2016-01-02 | 2 | 4 | | 2016-01-03 | 2 | 4 | +------------+------+------+
解决方法:
只需使用条件聚合:
SELECT DATE(date) as day,
SUM(author = 'Mary') AS Mary,
SUM(author = 'John') AS John
FROM chat
GROUP BY day
ORDER BY day ASC
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