Question

我有一个名为“results”的表，其中包含许多行和一个名为“timestamp”的列。

我想对每个月的行数进行分组。但对于他们中的每一个，我也想考虑前几个月的行。

所以，如果我有

Jun/13
Jun/13
Jun/13
Jul/13
Jul/13
Jul/13
Jul/13
Jul/13
Aug/13
Aug/13
Aug/13
Aug/13
Aug/13
Aug/13
Aug/13

结果将是

3  Jun/13
8  Jul/13
15 Aug/13

这就是我现在所拥有的，但不考虑前几个月。

SELECT DATE_FORMAT(FROM_UNIXTIME(timestamp), '%e %b %Y'), COUNT(1) FROM results
GROUP BY DATE_FORMAT(FROM_UNIXTIME(timestamp), '%e %b %Y');

Answer 1

我让这个工作：

SELECT month, count, @total:=@total+count AS total
FROM (SELECT @total:=0) t STRAIGHT_JOIN
  (SELECT DATE_FORMAT(FROM_UNIXTIME(timestamp), '%e %b %Y') AS month, COUNT(*) AS count
  FROM results GROUP BY EXTRACT(YEAR_MONTH FROM FROM_UNIXTIME(timestamp))) AS m;

+------------+-------+-------+
| month      | count | total |
+------------+-------+-------+
| 1 Jun 2013 |     3 |     3 |
| 1 Jul 2013 |     5 |     8 |
| 1 Aug 2013 |     7 |    15 |
+------------+-------+-------+

Answer 2

以下解决方案应该有效......

SELECT COUNT（1），DATE_FORMAT（timestamp，'％b /％y'）FROM FROM GROUP BY YEAR（时间戳），MONTH（时间戳）;

祝你好运

Talki

Answer 3

如果我理解正确，这应该有效;

SELECT COUNT(*) num, SUBSTR(ts, 1, 7) month
FROM results
JOIN (SELECT MAX(timestamp) ts FROM results
      GROUP BY YEAR(timestamp),Month(timestamp)) mm
  ON results.timestamp <= mm.ts
GROUP BY SUBSTR(ts, 1, 7);

An SQLfiddle to test with

按月对MySQL行进行分组，并将数据计算到每个月

3 个答案: