Question

如果我有一个名为[Part]的表，列[PartID]，[IDNumber]和[Length]和数据：

[PartID]  [IDNumber]  [Length]
1         Test1       50
2         Test1       60
3         Test2       50
4         Test3       70

如何只选择具有不同IDNumber的前2条记录？在搜索了一下后，我找不到能够做我想要的查询。我希望结果看起来像这样：

[PartID]  [IDNumber]  [Length]
1         Test1       50
3         Test2       50

我现在拥有的：

Select distinct top 2
        [PartID],
        [IDNumber],
        [Length]
from
    [Part]

澄清PartID实际上是GUID。我认为写出每个记录的GUID在我的示例数据中有点混乱。

Answer 1

SELECT DISTINCT TOP 2 PartId, IdNumber, Length
FROM
(   SELECT PartId, IdNumber, Length, ROW_NUMBER() over(partition by IdNumber order by Length) Orden
    FROM [Ayuda]
) A
WHERE A.Orden = 1
ORDER BY Length

Answer 2

SELECT TOP 2 b.* 
FROM   (SELECT idnumber, 
               MIN(partid) partid 
        FROM   part 
        GROUP  BY idnumber) a 
       JOIN part b 
         ON a.partid = b.partid  
ORDER  BY b.partid

Answer 3

您没有提到要为重复的IDNumber选择哪一行。在您的示例中，假设要使用Min PartID，您可以使用以下查询。需要稍微调整一下。

Select Top 2
    P.*
From
    [Part] P
Inner Join
    (
        Select
             [IDNumber]
            ,Min([PartID]) As MinPartID,
        From
            [Part]
        Group By
            [IDNumber]
    ) T
On
    P.PartID = T.MinPartID
    And
    P.IDNumber = T.IDNumber -- May not be needed is PartID is primary Key
Order By
     P.[PartID]
    ,P.[IDNumber]

SQL Select Distinct Top 2

3 个答案: