Question

使用SQL Server 2012

我有下表：

Id      Description 
6192    Salzburg
6193    Salzburg
6194    Salzburg
6196    Innsbruck
6197    Innsbruck
6198    Innsbruck
6199    Innsbruck
6201    Bregenz
6202    Bregenz
6203    Bregenz

我想在一个字符串中选择所有ID一起选择每个不同的“说明”：

    Description     Ids
    Salzburg        '6192,6193,6194'
    Innsbruck       '6196,6197,6198'

我在此网站[How to concatenate text from multiple rows into a single text string in SQL server?上看到了一些类似的代码，但出于我的目的（我不想使用XML路径！），我仍无法弄清楚。到目前为止，这是我尝试过的：

DECLARE @ids AS Nvarchar(MAX)
SELECT  @ids = COALESCE(@ids + ',', '') + CAST(t.Id AS nvarchar(5))
    FROM (SELECT tmp.Id FROM (SELECT id, [Description] FROM tblMasterPropValues WHERE IdCategory = 253 AND IsActive = 1) as tmp
        WHERE [Description] = tmp.[Description]) AS t
SELECT @ids
--SELECT DISTINCT [Description], @ids AS IDs FROM tblMasterPropValues WHERE IdCategory = 253 AND IsActive = 1 AND Id IN (@ids)

我实在无法解决问题，并且希望获得任何帮助。

Answer 1

您可以尝试使用STUFF()函数

SELECT description,  Ids = STUFF(
             (SELECT ',' + Id
              FROM tblMasterPropValues t1
              WHERE t1.description = t2.description
              FOR XML PATH (''))
             , 1, 1, '') from tblMasterPropValues t2
group by description;

Answer 2

因为FOR XML PATH()是正确的子句，所以，您可以执行以下操作：

SELECT DISTINCT v.description, STUFF(v1.ids, 1, 1, '''') + '''' 
FROM tblMasterPropValues v CROSS APPLY
     (SELECT ', '+ CAST(v1.Id AS VARCHAR(255))
      FROM tblMasterPropValues v1
      WHERE v1.description = v.description 
      FOR XML PATH('')
     ) v1(ids);

Answer 3

尝试一下：

DECLARE @Table TABLE(ID INT, Description VARCHAR(25))
INSERT INTO @Table
VALUES (6192,'Salzburg' )
,(6193,'Salzburg' )
,(6194,'Salzburg' )
,(6196,'Innsbruck')
,(6197,'Innsbruck')
,(6198,'Innsbruck')
,(6199,'Innsbruck')
,(6201,'Bregenz'     )
,(6202,'Bregenz'     )
,(6203,'Bregenz'     )

查询：

SELECT DISTINCT T2.Description, 
    SUBSTRING(
        (
            SELECT ','+CAST(T1.ID AS VARCHAR)  AS [text()]
            FROM @Table T1
            WHERE T1.Description = T2.Description
            ORDER BY T1.Description
            FOR XML PATH ('')
        ), 2, 1000) [Ids]
FROM @Table T2

结果：

Description Ids
Bregenz     6201,6202,6203
Innsbruck   6196,6197,6198,6199
Salzburg    6192,6193,6194

Answer 4

您还可以使用递归CTE来实现

DECLARE @tblMasterPropValues TABLE (Id INT, Description VARCHAR(20))
INSERT INTO @tblMasterPropValues VALUES
(6192 , 'Salzburg'),
(6193 , 'Salzburg'),
(6194 , 'Salzburg'),
(6196 , 'Innsbruck'),
(6197 , 'Innsbruck'),
(6198 , 'Innsbruck'),
(6199 , 'Innsbruck'),
(6201 , 'Bregenz'),
(6202 , 'Bregenz'),
(6203 , 'Bregenz')



;WITH Tbl AS 
(
    SELECT 
        *, 
        ROW_NUMBER() OVER(PARTITION BY Description ORDER BY Id) AS RN,
        COUNT(*) OVER(PARTITION BY Description) AS CNT
    FROM @tblMasterPropValues 
)
, Rcr AS (
    SELECT *, CAST(Id AS varchar(max)) Ids 
    FROM Tbl WHERE RN = 1
        UNION ALL
    SELECT T.*, Rcr.Ids + ',' + CAST(T.Id AS VARCHAR(10)) Ids 
    FROM Rcr 
        INNER JOIN Tbl T ON T.RN = Rcr.RN + 1 and Rcr.Description = T.Description
)
SELECT RN, Description, Ids FROM Rcr 
WHERE RN = CNT

结果：

Description          Ids
-------------------- -----------------------
Salzburg             6192,6193,6194
Innsbruck            6196,6197,6198,6199
Bregenz              6201,6202,6203

结合多行与COALESCE

4 个答案: