在单个SQL查询中，不使用FOR XML子句。 公共表表达式用于递归地连接结果。

-- rank locations by incrementing lexicographical order
WITH RankedLocations AS (
  SELECT
    VehicleID,
    City,
    ROW_NUMBER() OVER (
        PARTITION BY VehicleID 
        ORDER BY City
    ) Rank
  FROM
    Locations
),
-- concatenate locations using a recursive query
-- (Common Table Expression)
Concatenations AS (
  -- for each vehicle, select the first location
  SELECT
    VehicleID,
    CONVERT(nvarchar(MAX), City) Cities,
    Rank
  FROM
    RankedLocations
  WHERE
    Rank = 1

  -- then incrementally concatenate with the next location
  -- this will return intermediate concatenations that will be 
  -- filtered out later on
  UNION ALL

  SELECT
    c.VehicleID,
    (c.Cities + ', ' + l.City) Cities,
    l.Rank
  FROM
    Concatenations c -- this is a recursion!
    INNER JOIN RankedLocations l ON
        l.VehicleID = c.VehicleID 
        AND l.Rank = c.Rank + 1
),
-- rank concatenation results by decrementing length 
-- (rank 1 will always be for the longest concatenation)
RankedConcatenations AS (
  SELECT
    VehicleID,
    Cities,
    ROW_NUMBER() OVER (
        PARTITION BY VehicleID 
        ORDER BY Rank DESC
    ) Rank
  FROM 
    Concatenations
)
-- main query
SELECT
  v.VehicleID,
  v.Name,
  c.Cities
FROM
  Vehicles v
  INNER JOIN RankedConcatenations c ON 
    c.VehicleID = v.VehicleID 
    AND c.Rank = 1

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。

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。

SQL Server 2005

SELECT Stuff(
  (SELECT N', ' + Name FROM Names FOR XML PATH(''),TYPE)
  .value('text()[1]','nvarchar(max)'),1,2,N'')

SQL Server 2016

你可以使用FOR JSON语法

即。

SELECT per.ID,
Emails = JSON_VALUE(
   REPLACE(
     (SELECT _ = em.Email FROM Email em WHERE em.Person = per.ID FOR JSON PATH)
    ,'"},{"_":"',', '),'$[0]._'
) 
FROM Person per

结果就会变成

Id  Emails
1   abc@gmail.com
2   NULL
3   def@gmail.com, xyz@gmail.com

即使数据包含无效的XML字符，这也可以工作

的 '"},{"":"' 你是安全的,因为如果数据包含吗 '"},{"":"', 它将逃到 "},{\"_\":\"

你可以用任何字符串分隔符替换'，'

在SQL Server 2017, Azure SQL数据库

您可以使用新的STRING_AGG函数

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

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]