我们可以使用RECUSRSIVITY、WITH CTE、union ALL，如下所示

declare @mytable as table(id int identity(1,1), str nvarchar(100))
insert into @mytable values('Peter'),('Paul'),('Mary')

declare @myresult as table(id int,str nvarchar(max),ind int, R# int)

;with cte as(select id,cast(str as nvarchar(100)) as str, cast(0 as int) ind from @mytable
union all
select t2.id,cast(t1.str+',' +t2.str as nvarchar(100)) ,t1.ind+1 from cte t1 inner join @mytable t2 on t2.id=t1.id+1)
insert into @myresult select *,row_number() over(order by ind) R# from cte

select top 1 str from @myresult order by R# desc

你需要创建一个变量来保存你的最终结果并选择它，就像这样。

最简单的解决方案

DECLARE @char VARCHAR(MAX);

SELECT @char = COALESCE(@char + ', ' + [column], [column]) 
FROM [table];

PRINT @char;

如果您使用的是SQL Server 2017或Azure，请参阅Mathieu Renda的回答。

当我试图连接两个具有一对多关系的表时，我也遇到了类似的问题。在SQL2005中，我发现XMLPATH方法可以非常容易地处理行的连接。

如果有一个名为STUDENTS的表

SubjectID       StudentName
----------      -------------
1               Mary
1               John
1               Sam
2               Alaina
2               Edward

我期望的结果是：

SubjectID       StudentName
----------      -------------
1               Mary, John, Sam
2               Alaina, Edward

我使用了以下T-SQL：

SELECT Main.SubjectID,
       LEFT(Main.Students,Len(Main.Students)-1) As "Students"
FROM
    (
        SELECT DISTINCT ST2.SubjectID, 
            (
                SELECT ST1.StudentName + ',' AS [text()]
                FROM dbo.Students ST1
                WHERE ST1.SubjectID = ST2.SubjectID
                ORDER BY ST1.SubjectID
                FOR XML PATH (''), TYPE
            ).value('text()[1]','nvarchar(max)') [Students]
        FROM dbo.Students ST2
    ) [Main]

如果您可以在开头插入逗号并使用子字符串跳过第一个逗号，那么您可以以更紧凑的方式执行相同的操作，这样就不需要执行子查询：

SELECT DISTINCT ST2.SubjectID, 
    SUBSTRING(
        (
            SELECT ','+ST1.StudentName  AS [text()]
            FROM dbo.Students ST1
            WHERE ST1.SubjectID = ST2.SubjectID
            ORDER BY ST1.SubjectID
            FOR XML PATH (''), TYPE
        ).value('text()[1]','nvarchar(max)'), 2, 1000) [Students]
FROM dbo.Students ST2

在Chris Shaffer的回答之上：

如果您的数据可能会重复，例如

Tom
Ali
John
Ali
Tom
Mike

而不是汤姆、阿里、约翰、阿里、汤姆、迈克

您可以使用DISTINCT避免重复，并让Tom、Ali、John、Mike：

DECLARE @Names VARCHAR(8000)
SELECT DISTINCT @Names = COALESCE(@Names + ',', '') + Name
FROM People
WHERE Name IS NOT NULL
SELECT @Names

提出了递归CTE解决方案，但没有提供代码。下面的代码是递归CTE的示例。

请注意，虽然结果与问题相符，但数据与给定的描述并不完全相符，因为我假设您确实希望对行组（而不是表中的所有行）执行此操作。将其更改为与表中的所有行相匹配是读者的练习。

;WITH basetable AS (
    SELECT
        id,
        CAST(name AS VARCHAR(MAX)) name,
        ROW_NUMBER() OVER (Partition BY id ORDER BY seq) rw,
        COUNT(*) OVER (Partition BY id) recs
    FROM (VALUES
        (1, 'Johnny', 1),
        (1, 'M', 2),
        (2, 'Bill', 1),
        (2, 'S.', 4),
        (2, 'Preston', 5),
        (2, 'Esq.', 6),
        (3, 'Ted', 1),
        (3, 'Theodore', 2),
        (3, 'Logan', 3),
        (4, 'Peter', 1),
        (4, 'Paul', 2),
        (4, 'Mary', 3)
    ) g (id, name, seq)
),
rCTE AS (
    SELECT recs, id, name, rw
    FROM basetable
    WHERE rw = 1

    UNION ALL

    SELECT b.recs, r.ID, r.name +', '+ b.name name, r.rw + 1
    FROM basetable b
    INNER JOIN rCTE r ON b.id = r.id AND b.rw = r.rw + 1
)
SELECT name
FROM rCTE
WHERE recs = rw AND ID=4
OPTION (MAXRECURSION 101)

此方法仅适用于Teradata Aster数据库，因为它使用NPATH函数。

再次，我们有桌上学生

SubjectID       StudentName
----------      -------------
1               Mary
1               John
1               Sam
2               Alaina
2               Edward

然后使用NPATH，只需一次SELECT：

SELECT * FROM npath(
  ON Students
  PARTITION BY SubjectID
  ORDER BY StudentName
  MODE(nonoverlapping)
  PATTERN('A*')
  SYMBOLS(
    'true' as A
  )
  RESULT(
    FIRST(SubjectID of A) as SubjectID,
    ACCUMULATE(StudentName of A) as StudentName
  )
);

结果：

SubjectID       StudentName
----------      -------------
1               [John, Mary, Sam]
2               [Alaina, Edward]