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Mysql-是否可以对之前的行值进行累加

[英]Mysql - Is it possible to cumlatively sum the previous row values

I have been trying this for a while but no success. 我已经尝试了一段时间,但没有成功。

My database is something like this 我的数据库是这样的

-----------------------------------------------------
DEVICE_HASH | DEVICE_NAME | DEVICE_VERISON | DATE
-----------------------------------------------------
122138223823| crespo      | 1.1            | 12/01/13
122138213823| jewell      | 1.2            | 12/01/13
122138223823| crespo      | 1.1            | 13/01/13
122138263823| crespo      | 1.1            | 13/01/13
122138283823| blade       | 1.2            | 13/01/13
122138293823| crespo      | 1.1            | 14/01/13
-----------------------------------------------------

To get the count on devices downloaded per day i executed the below query which worked fine. 为了获得每天下载设备的数量,我执行了以下查询,效果很好。

SELECT DATE_FORMAT(DATE, '%Y-%m-%d') as 'date', DEVICE_NAME as 'device' ,COUNT(*) as 'count' FROM table_name GROUP BY DATE_FORMAT(DATE, '%Y-%m-%d'), DEVICE_NAME

Eg: I got crespo count as 1 on 12/01/13, 2 on 13/01/13 and 1 on 14/01/13 例如:我在12/01/13上的crespo计数为1,在13/01/13上的计数为2,在14/01/13上的计数为1

Now i want to get the running count of devices ie add previous day value to current day download and go on until the end. 现在,我想获取设备的运行计数,即将前一天的值添加到当前一天的下载中,并继续进行到最后。

Eg: crespo count should go as 1 on 12/01/13, 3 on 13/01/13 and 4 on 14/01/13 例如:crespo计数应在12/01/13时为1,在13/01/13时为3,在14/01/13时为4

Is this possible in mysql ? 在mysql中这可能吗?

I think this is what you're looking for: 我认为这是您要寻找的:

SELECT
  @runningTotal:=IF(@prevDevice=Device_Name,@runningTotal+cnt,cnt) rt,
  dt, Device_Name, cnt,
  @prevDevice:=Device_Name
FROM (
  SELECT 
    DATE_FORMAT(DATE, '%Y-%m-%d') as dt, 
    DEVICE_NAME ,
    COUNT(*) as cnt
  FROM table_name 
  GROUP BY DATE_FORMAT(DATE, '%Y-%m-%d'), DEVICE_NAME
  ORDER BY DEVICE_NAME, 2
) t
  JOIN (SELECT @runningTotal:=0) r

And the sample fiddle: http://sqlfiddle.com/#!2/e3525/1 和示例提琴: http ://sqlfiddle.com/#!2/ e3525/1

Which produces the following results: 产生以下结果:

RT  DT          DEVICE_NAME     CNT 
1   2013-01-13  blade           1   
1   2014-01-13  crespo          1   
3   2013-01-13  crespo          2   
4   2012-01-13  crespo          1   
1   2012-01-13  jewell          1   

try this , 尝试这个 ,
you need extra variable to keep check of count , 你需要额外的变量来保持计数

fiddle example 小提琴的例子

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