0
Date                Time        Method          User                   App
2013/5/24        19:39:33        PUT        advantage_user        adv_cat_arch
2013/5/24        19:36:04        SETACL        advantage_user        adv_cat_arch
2013/5/24        19:36:11        PUT        advantage_user        adv_cat_arch
2013/5/24        19:36:12        SETACL        advantage_user        adv_cat_arch
2013/5/24        19:36:19        PUT        advantage_user        adv_cat_arch
2013/5/24        19:36:19        SETACL        advantage_user        adv_cat_arch
2013/5/24        19:36:27        PUT        advantage_user        adv_cat_arch

我想计算每天的每种方法,例如,

Date        Method(PUT)    Method(SETACL)  Method(Get)
2013/5/24       5                2             3
2013/5/25       2                1             5

这是我的 sql 查询:

SELECT Fas_ops_metrics_test4_jun.Date as Date, 
       Fas_ops_metrics_test4_jun.Method as Method
FROM Fas_ops_metrics_test4_jun
GROUP BY Fas_ops_metrics_test4_jun.Method
ORDER BY Fas_ops_metrics_test4_jun.Method;

任何人都可以帮助我吗?谢谢你

4

2 回答 2

1

如果我对您的理解正确,您可以使用COUNTand执行此操作CASE

SELECT Date, 
    COUNT(CASE WHEN Method = 'PUT' THEN 1 END) 'Method(PUT)',
    COUNT(CASE WHEN Method = 'SETACL' THEN 1 END) 'Method(SETACL)',
    COUNT(CASE WHEN Method = 'GET' THEN 1 END) 'Method(GET)'
FROM YourTable
GROUP BY Date

编辑,假设您使用的是 MS Access,请使用IIF以下命令代替,CASE也许更容易SUM

SELECT Date, 
    SUM(IIF(Method = 'PUT', 1, 0)) as 'Method(PUT)',
    SUM(IIF(Method = 'SETACL', 1, 0)) as 'Method(SETACL)',
    SUM(IIF(Method = 'GET', 1, 0)) as 'Method(GET)'
FROM YourTable
GROUP BY Date
于 2013-07-05T22:38:29.890 回答
0
Select count (method) , method
from yourTable 
Where Date = '2013/5/24'
group by method

这将为您提供两列,每个方法和方法的计数。

于 2013-07-05T22:35:19.977 回答