为了说明这一点,假设我有如下两个表:

VehicleID Name
1         Chuck
2         Larry

LocationID VehicleID City
1          1         New York
2          1         Seattle
3          1         Vancouver
4          2         Los Angeles
5          2         Houston

我想写一个查询返回以下结果:

VehicleID Name    Locations
1         Chuck   New York, Seattle, Vancouver
2         Larry   Los Angeles, Houston

我知道这可以使用服务器端游标完成,即:

DECLARE @VehicleID int
DECLARE @VehicleName varchar(100)
DECLARE @LocationCity varchar(100)
DECLARE @Locations varchar(4000)
DECLARE @Results TABLE
(
  VehicleID int
  Name varchar(100)
  Locations varchar(4000)
)

DECLARE VehiclesCursor CURSOR FOR
SELECT
  [VehicleID]
, [Name]
FROM [Vehicles]

OPEN VehiclesCursor

FETCH NEXT FROM VehiclesCursor INTO
  @VehicleID
, @VehicleName
WHILE @@FETCH_STATUS = 0
BEGIN

  SET @Locations = ''

  DECLARE LocationsCursor CURSOR FOR
  SELECT
    [City]
  FROM [Locations]
  WHERE [VehicleID] = @VehicleID

  OPEN LocationsCursor

  FETCH NEXT FROM LocationsCursor INTO
    @LocationCity
  WHILE @@FETCH_STATUS = 0
  BEGIN
    SET @Locations = @Locations + @LocationCity

    FETCH NEXT FROM LocationsCursor INTO
      @LocationCity
  END
  CLOSE LocationsCursor
  DEALLOCATE LocationsCursor

  INSERT INTO @Results (VehicleID, Name, Locations) SELECT @VehicleID, @Name, @Locations

END     
CLOSE VehiclesCursor
DEALLOCATE VehiclesCursor

SELECT * FROM @Results

然而,正如您所看到的,这需要大量的代码。我想要的是一个泛型函数,允许我做这样的事情:

SELECT VehicleID
     , Name
     , JOIN(SELECT City FROM Locations WHERE VehicleID = Vehicles.VehicleID, ', ') AS Locations
FROM Vehicles

这可能吗?或者类似的东西?


当前回答

试试这个查询

SELECT v.VehicleId, v.Name, ll.LocationList
FROM Vehicles v 
LEFT JOIN 
    (SELECT 
     DISTINCT
        VehicleId,
        REPLACE(
            REPLACE(
                REPLACE(
                    (
                        SELECT City as c 
                        FROM Locations x 
                        WHERE x.VehicleID = l.VehicleID FOR XML PATH('')
                    ),    
                    '</c><c>',', '
                 ),
             '<c>',''
            ),
        '</c>', ''
        ) AS LocationList
    FROM Locations l
) ll ON ll.VehicleId = v.VehicleId

其他回答

我通过创建以下函数找到了一个解决方案:

CREATE FUNCTION [dbo].[JoinTexts]
(
  @delimiter VARCHAR(20) ,
  @whereClause VARCHAR(1)
)
RETURNS VARCHAR(MAX)
AS 
BEGIN
    DECLARE @Texts VARCHAR(MAX)

    SELECT  @Texts = COALESCE(@Texts + @delimiter, '') + T.Texto
    FROM    SomeTable AS T
    WHERE   T.SomeOtherColumn = @whereClause

    RETURN @Texts
END
GO

用法:

SELECT dbo.JoinTexts(' , ', 'Y')

妈妈的答案对我没用,所以我对答案做了一些修改,让它起作用。希望这能帮助到一些人。 使用SQL Server 2012:

SELECT [VehicleID]
     , [Name]
     , STUFF((SELECT DISTINCT ',' + CONVERT(VARCHAR,City) 
         FROM [Location] 
         WHERE (VehicleID = Vehicle.VehicleID) 
         FOR XML PATH ('')), 1, 2, '') AS Locations
FROM [Vehicle]

从我所看到的FOR XML(如前所述)是唯一的方法,如果你还想选择其他列(我猜大多数会),就像OP所做的那样。 使用合并(@var……不允许包含其他列。

更新: 感谢programmingsolutions.net,有一种方法可以删除“尾随”逗号。 通过使它成为一个前导逗号,并使用MSSQL的STUFF函数,你可以将第一个字符(前导逗号)替换为一个空字符串,如下所示:

stuff(
    (select ',' + Column 
     from Table
         inner where inner.Id = outer.Id 
     for xml path('')
), 1,1,'') as Values

我不相信有一种方法可以在一个查询中完成它,但你可以用一个临时变量来玩这样的技巧:

declare @s varchar(max)
set @s = ''
select @s = @s + City + ',' from Locations

select @s

这绝对比在游标上移动要少的代码,而且可能更有效率。

如果你正在运行SQL Server 2005,你可以编写一个自定义CLR聚合函数来处理这个问题。

c#版本:

using System;
using System.Data;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using System.Text;
using Microsoft.SqlServer.Server;
[Serializable]
[Microsoft.SqlServer.Server.SqlUserDefinedAggregate(Format.UserDefined,MaxByteSize=8000)]
public class CSV:IBinarySerialize
{
    private StringBuilder Result;
    public void Init() {
        this.Result = new StringBuilder();
    }

    public void Accumulate(SqlString Value) {
        if (Value.IsNull) return;
        this.Result.Append(Value.Value).Append(",");
    }
    public void Merge(CSV Group) {
        this.Result.Append(Group.Result);
    }
    public SqlString Terminate() {
        return new SqlString(this.Result.ToString());
    }
    public void Read(System.IO.BinaryReader r) {
        this.Result = new StringBuilder(r.ReadString());
    }
    public void Write(System.IO.BinaryWriter w) {
        w.Write(this.Result.ToString());
    }
}