按月对MySQL行进行分组并计数直到每个月

Question

我有一个名为“results”的表，其中包含大量行和名为“timestamp”的列。

我想将每个月的行数分组。但对于其中的每一个，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 

其结果将是

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

这就是我现在所拥有的，但它并没有考虑到前几个月。

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

我得到这个工作：

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

以下解决方案应该是工作...

SELECT COUNT（1），DATE_FORMAT（时间戳， '％B /％Y'）FROM结果 GROUP BY YEAR（时间戳），月（时间戳）;

好运

Talki

Answer 3

如果我理解正确的话，这应该工作;

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。