这里有一个日期的例子。

create table t1(created date not null);

insert 
  into t1(created) values (date '2011-01-09')
                         ,(date '2011-01-10')
                         ,(date '2011-01-11')
                         ,(date '2011-01-12')
                         ,(date '2011-01-13')
                         ,(date '2011-01-14')
                         ,(date '2011-01-15')
                         ,(date '2011-01-16')
                         ,(date '2011-01-17')
                         ,(date '2011-01-18')
                         ,(date '2011-01-19')
                         ,(date '2011-01-20');

select floor(datediff(now(), created) / 3) * 3 as days_ago
      ,min(created)
      ,max(created)
      ,count(*)
  from t1
 group 
    by floor(datediff(now(), created) / 3);

+----------+--------------+--------------+----------+
| days_ago | min(created) | max(created) | count(*) |
+----------+--------------+--------------+----------+
|        0 | 2011-01-18   | 2011-01-20   |        3 |
|        3 | 2011-01-15   | 2011-01-17   |        3 |
|        6 | 2011-01-12   | 2011-01-14   |        3 |
|        9 | 2011-01-09   | 2011-01-11   |        3 |
+----------+--------------+--------------+----------+
4 rows in set (0.00 sec)

你可以这样做
SELECT COUNT(*) FROM table GROUP BY FLOOR(created / 3)
我想是的
虽然如果created是一个日期字段，你必须做更多的调整，使它成为一个数字值，这样才能工作。

谢谢@Ronnis，我用你的例子，终于解决了我的问题。
还有一个小错误我发现，在例子中，我添加了一行

insert into t1(created) values (date '2011-01-21')

现在我得到：

+----------+--------------+--------------+----------+
| days_ago | min(created) | max(created) | count(*) |
+----------+--------------+--------------+----------+
|     1986 | 2011-01-20   | 2011-01-21   |        2 |
|     1989 | 2011-01-17   | 2011-01-19   |        3 |
|     1992 | 2011-01-14   | 2011-01-16   |        3 |
|     1995 | 2011-01-11   | 2011-01-13   |        3 |
|     1998 | 2011-01-09   | 2011-01-10   |        2 |
+----------+--------------+--------------+----------+

正如你所看到的，日子分成了2，3，3，3，2，这不是我所期望的。
我将SQL更改为

select floor(datediff(created, '2011-01-09') / 3) * 3 as days_before
      ,min(created)
      ,max(created)
      ,count(*)
from t1
group by floor(datediff(created, '2011-01-09') / 3);

只需获取从created到start date的datediff，现在我得到：

+-------------+--------------+--------------+----------+
| days_before | min(created) | max(created) | count(*) |
+-------------+--------------+--------------+----------+
|           0 | 2011-01-09   | 2011-01-11   |        3 |
|           3 | 2011-01-12   | 2011-01-14   |        3 |
|           6 | 2011-01-15   | 2011-01-17   |        3 |
|           9 | 2011-01-18   | 2011-01-20   |        3 |
|          12 | 2011-01-21   | 2011-01-21   |        1 |
+-------------+--------------+--------------+----------+

那可能更合适。