2016-08-17 35 views
0

我正在上传文件上传成功的文件上传项目。基于相同的柱头在mysql中分组php

我现在正在显示其用于存储上传的文件细节我上传的表是这样的:

+--------------------+--------------------+------------------+-----------+ 
| projectId  | filename   | projectDetails |ProjectName | 
+--------------------+--------------------+------------------+------------+ 
| pid1    | admin.zip   | This Is A Test |myProject | 
| pid1    | admin1.zip   | This Is A Test |myProject | 
| pid2    | admin2.zip   | This Is A Test2 |myProject2 | 
| pid2    | admin3.zip   | This Is A Test2 |myProject2 | 
| pid1    | admin4.zip   | This Is A Test |myProject | 
+--------------------+--------------------+------------------+------------+ 

可我知道,我可以得到这样的

+--------------------+--------------------+------------------+-----------+ 
| projectId  | filename   | projectDetails |ProjectName | 
+--------------------+--------------------+------------------+------------+ 
| pid1    | admin.zip   | This Is A Test |myProject | 
|     | admin1.zip   |     |   | 
|     | admin4.zip   |     |   | 
+--------------------+--------------------+------------------+------------+ 
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这是关于数据的表示。在'MySQL'结尾没有任何事情要做。 – 1000111

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感谢重播 –

回答

-1

输出什么关于使用此查询:

select projectId, GROUP_CONCAT(distinct filename) as filename, projectDetails, ProjectName 
from yourtable 
group by pid1 
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是的,它的工作,我得到了什么,我正在寻找你是awsome –

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@shahidkhan请注意,这将返回任意projectDetails和ProjectName。 – Strawberry

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我该如何避免这种情况 –