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

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')

注意,Matt的代码将导致在字符串的末尾增加一个逗号;使用COALESCE(或ISNULL),如Lance的链接中所示,使用了类似的方法,但没有给你留下一个额外的逗号。为了完整起见,下面是兰斯在sqlteam.com上的相关代码:

DECLARE @EmployeeList varchar(100)
SELECT @EmployeeList = COALESCE(@EmployeeList + ', ', '') + 
    CAST(EmpUniqueID AS varchar(5))
FROM SalesCallsEmployees
WHERE SalCal_UniqueID = 1

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

如果你使用的是SQL Server 2005,你可以使用FOR XML PATH命令。

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

这比使用光标容易得多,而且似乎工作得相当好。

更新

对于那些在新版本的SQL Server中仍然使用这种方法的人来说,还有另一种方法,它更简单,性能更好 自SQL Server 2017以来已经可用的STRING_AGG方法。

SELECT  [VehicleID]
       ,[Name]
       ,(SELECT STRING_AGG([City], ', ')
         FROM [Location]
         WHERE VehicleID = V.VehicleID) AS Locations
FROM   [Vehicle] V

这还允许将不同的分隔符指定为第二个参数,比前一种方法提供了更多的灵活性。