SQL：为表中的所有值集显示不同的ID [英] SQL:Display distinct ids for all the set of values from table

问题描述

我遇到一个问题，在执行查询后，我得到这样的结果

I have a problem where after executing a query i'm getting a result like this

DevID   Difference      
-----------------
99       5        
99       10   
99       5        
99       4 
12       8        
12       9 
12       5        
12       6 

我不希望重复的ID ，
我应该只能显示一个id。
这可以通过使用distinct轻松实现，但是问题是我还需要显示差异列。我不介意diff中包含哪个值，但是99的值之一可以在那里，但基本上我只需要一个id值。
预期结果是这样的。

i dont want the duplicate ids, I should be able to display only one id. This could be easily achieved by using distinct however the problem is i also need to display the Difference column. I'm not bothered which value comes in diff but either one of the values for 99 can come there but basically i just need one value for id. Expected result is something like this.

DevID   Difference      
-----------------
99       5        
12       8  

如果它有助于此问题是以下问题的继续
属于同一组但存储在两行中的值的差异

If it helps this question is continuation of the following question Difference of values that belong to same group but stored in two rows

推荐答案

您可以将 GROUP BY 与 MAX 或 MIN 一起使用，或者

You can use GROUP BY along with MAX or MIN or any other aggregation function to get a result as you expected.

SELECT DevID, MAX(Difference) 
FROM Yourtablename
GROUP BY DevID

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