PHP-MySQL：每个日期的分组出现次数

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.

我有一个聊天表,看起来像：

+---------+----------------+---------------------+-----------------+
| 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 |
+------------+------+------+

我有义务在COUNT语句中执行子查询吗？

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 |
+------------+------+------+

任何帮助或相关主题的链接将不胜感激.
祝你有美好的一天

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