Question

我有这张桌子

Username    Client Name         Date        Time            Published App
1stLineS    GS-HPS7             03/08/2015   9:37:44.12     PB Service Desktop
1stLineS    GS-HPS7             09/07/2015  11:08:42.31     Citrix Delivery Services Console
1stLineS    GARION-WIN8         29/07/2015  15:34:19.85     Mencap@Work2-1
aabidam     009733-DSK          03/08/2015   9:41:56.25     Epicor
aabidam     009733-DSK          03/08/2015  11:53:59.78     Mencap@Work2-1
aabidam     010175-LAP          06/08/2015   9:23:15.67     Mencap Desktop with Acrobat
aabidam     NC-CITRIXDESKAA     06/08/2015  12:54:23.51     Powerplan
aabidam     WT008064c025aa      13/08/2015  13:44:39.34     Thin Client Desktop
aarong      NCIT-008888-DSK     25/07/2015   7:13:30.47     Mencap @ Work Desktop
AarthiT     HTML-5190-7294      21/07/2015  11:23:03.42     Mencap@Work2-1
AbbeyY      005625-RMS          14/07/2015  14:22:17.37     Mencap@Work2-1
abbiemc     008700-LAP          03/08/2015  12:31:35.12     Mencap@Work2-1
abbyf       004503-DSK          08/07/2015   9:10:30.71     Mencap@Work2-1
abbygailf   010030-LAP          03/08/2015  11:32:34.53     Mencap@Work2-1
Abbyh       ABBY-PC             03/08/2015  13:59:25.21     GLH Contractors Web Portal
abdio       009899-IT           10/07/2015  18:07:49.17     PB Service Desktop
abdio       MEMBER-PC           11/08/2015  19:37:49.94     RDP NC-BACKUP01
abdouD      005240-LAP          03/08/2015  11:43:23.43     Mencap@Work2-1
abdulm      008617-DSK          03/08/2015   8:44:32.54     Mencap@Work2-1
AbigailE    HP81017347825       02/08/2015  16:07:48.10     Mencap@Work2-1
AbigailJ    009178-DSK          03/08/2015   9:07:03.91     Mencap@Work2-1
AbigailJ    009178-DSK          07/07/2015  10:58:33.74     Mencap Desktop with Acrobat
AbigailJ    009178-DSK          07/07/2015  10:59:33.74     Epicor

我需要让用户回到Mencap @ Work2-1和桌面发布的应用程序＆＃34;这是一组特定的应用程序，我会回复此用户列表

1stLineS
aabidam
AbigailJ

我没有得到具体结果......

我已尝试过以下查询并返回0

$sql = "SELECT *
FROM tbl_name
WHERE publishedapp LIKE \"%Mencap@Work2-1%\"
AND publishedapp LIKE \"%Desktop%\"
AND publishedapp  LIKE \"%RDP%\"
GROUP BY username


$sql = "SELECT DISTINCT username, publishedapp
FROM tbl_name
WHERE publishedapp = \"Mencap@Work2-1\"
  AND publishedapp   NOT LIKE \"%Desktop%\"
  AND publishedapp   NOT LIKE \"%RDP%\"
GROUP BY username
HAVING COUNT(DISTINCT publishedapp) > 1
";

我尝试了更多的变化，但说实话，我不知道我在做什么，我有点脑死亡

Answer 1

您需要having子句和group by：

SELECT username
FROM tbl_name
GROUP BY username
HAVING SUM(publishedapp LIKE '%Mencap@Work2-1%') > 0 AND
       SUM(publishedapp LIKE '%Desktop%') AND
       SUM(publishedapp LIKE '%RDP%');

Answer 2

这解决了我的问题

SELECT DISTINCT username,
GROUP_CONCAT(DISTINCT publishedapp SEPARATOR \"<br>\") AS publishedapp
FROM tbl_name
WHERE publishedapp LIKE \"%Mencap@Work2-1%\"
OR publishedapp LIKE \"%@%\"
OR publishedapp LIKE \"%Desktop%\"
OR publishedapp LIKE \"%RDP%\"
GROUP BY username
HAVING COUNT(DISTINCT publishedapp) > 1 AND
SUM(DISTINCT publishedapp LIKE \"%Mencap@Work2-1%\") > 0

不仅返回了用户列表，而且返回了一个包含所用应用列表的字符串，例如

1stLineS
  Mencap@work2-1
  Citrix Delivery Services Console
aabidam
  Epicor
  Mencap@Work2-1
  Powerplan
AbigailJ
  Mencap@Work2-1
  Epicor

SQL Query，选择distinct，FROM，where或者

2 个答案: