Question

我有以下mysql表Test_Results

| ID | Test_ID | Device_OS| Status | Date    |
| 1  | 1       | Android  | passed | 2015/10 |
| 2  | 1       | Android  | failed | 2015/10 |
| 3  | 15      | iOS      | passed | 2015/11 |
| 4  | 20      | Android  | passed | 2015/11 |
| 5  | 27      | Android  | passed | 2015/11 |
| 6  | 10      | iOS      | failed | 2015/12 |
| 7  | 5       | Android  | passed | 2015/12 |

我需要按月计算每个操作系统和组的唯一Test_ID的数量。我写了一个查询来计算唯一测试的一般数量：

SELECT Run_Date, COUNT(DISTINCT Test_ID) FROM Test_Results
GROUP BY MONTH(Run_Date)
ORDER BY Run_Date DESC;

但我无法弄清楚如何打破Deivce_OS。所以结果会看起来像这样：

Run_Date    COUNT Android iOS
12/10/2015  3650  3650    2500
11/5/2015   1909  1909    1900
10/1/2015   1667  1667    1325

谢谢！

Answer 1

首先，你不应该按月分组，除非你真的，真的打算这样做。其次，你应该习惯只使用SELECT中GROUP BY中的表达式。

第三，我认为你只需要条件聚合，如下所示：

SELECT YEAR(Run_Date), MONTH(Run_Date), COUNT(DISTINCT Test_ID),
       SUM(Device_OS = 'Android') as Android, SUM(Device_OS = 'iOS') as iOS
FROM Test_Results
GROUP BY YEAR(Run_Date), MONTH(RUn_Date)
ORDER BY YEAR(Run_Date) DESC, MONTH(RUn_Date) DESC;

编辑：

我注意到您希望每月为操作系统进行唯一测试。为此，请以条件方式使用COUNT(DISTINCT)：

SELECT YEAR(Run_Date), MONTH(Run_Date), COUNT(DISTINCT Test_ID),
       COUNT(DISTINCT CASE WHEN Device_OS = 'Android' THEN Test_ID END) as Android,
       COUNT(DISTINCT CASE WHEN Device_OS = 'iOS' THEN Test_ID END) as iOS
FROM Test_Results
GROUP BY YEAR(Run_Date), MONTH(RUn_Date)
ORDER BY YEAR(Run_Date) DESC, MONTH(RUn_Date) DESC;

MySQL：计算if和Group By

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