小编典典

MySQL选择MAX(日期时间)不返回最大值

sql

表格示例:

id   computer  app      version     build    date
---|---------|------|------------|-------|---------
1  |  aaaa1  | app1 |   1.0.0    |   1   | 2013-11-11 09:51:07
2  |  aaaa1  | app2 |   2.0.0    |   2   | 2013-11-12 09:51:07
5  |  xxxx2  | app1 |   1.0.0    |   1   | 2013-11-13 09:51:07
3  |  cccc3  | app2 |   3.1.0    |   1   | 2013-11-14 09:51:07
4  |  xxxx2  | app1 |   1.0.0    |   2   | 2013-11-15 09:51:07
5  |  cccc3  | app2 |   3.1.1    |   3   | 2013-11-16 09:51:07
6  |  xxxx2  | app1 |   1.0.2    |   1   | 2013-11-17 09:51:07
7  |  aaaa1  | app1 |   1.0.2    |   3   | 2013-11-18 09:51:07

所需的输出(不是确切的格式或上市顺序),每台计算机上每个应用程序的最新安装:

7. aaaa1 - app1 - 1.0.2 - 3 - 2013-11-18 09:51:07
2. aaaa1 - app2 - 2.0.0 - 2 - 2013-11-12 09:51:07
6. xxxx2 - app1 - 1.0.2 - 1 - 2013-11-17 09:51:07
5. cccc3 - app2 - 3.1.1 - 3 - 2013-11-16 09:51:07

我的SQL语句:

SELECT 
        id,
        computer, 
        app, 
        version, 
        build, 
        MAX(date) AS installed
    FROM 
        data 
    WHERE 
        placement = 'xxx'
    GROUP BY 
        app, computer
    ;

这给了我:

1. aaaa1 - app1 - 1.0.0 - 1 - 2013-11-11 09:51:07

并不是

7. aaaa1 - app1 - 1.0.2 - 3 - 2013-11-18 09:51:07

如我所料。

如果我 选择MAX(date),而没有其他选择,则MAX(date)有效。但是,那时我没有任何数据可以使用(仅是最新日期)。

SELECT 
        MAX(date) AS installed

我不是SQL忍者,所以我很快就会因此而挠头。


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2021-05-05

共1个答案

小编典典

尝试这样:

SELECT d.id, d.computer, d.app, d.version, d.build, a.installed
FROM data d
INNER JOIN (
  SELECT computer, app, max(DATE) AS installed
  FROM data
  GROUP BY computer, app
  ) a ON a.computer = d.computer AND a.app = d.app
WHERE placement = 'xxx'

内部查询是为您获取每对计算机和应用程序的max(date)最大值,然后您只需与之结合即可获取其余信息。

2021-05-05