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

Oracle 11g Release 2支持LISTAGG功能。此处的文档。

COLUMN employees FORMAT A50

SELECT deptno, LISTAGG(ename, ',') WITHIN GROUP (ORDER BY ename) AS employees
FROM   emp
GROUP BY deptno;

    DEPTNO EMPLOYEES
---------- --------------------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

警告

如果生成的字符串可能超过4000个字符，请小心执行此函数。它将抛出异常。如果是这种情况，那么您需要处理异常或滚动自己的函数，以防止连接的字符串超过4000个字符。

一个现成的解决方案，没有额外的逗号：

select substring(
        (select ', '+Name AS 'data()' from Names for xml path(''))
       ,3, 255) as "MyList"

空列表将导致NULL值。通常，您会将列表插入到表列或程序变量中：根据需要调整最大长度255。

（迪瓦卡尔和延斯·弗兰森提供了很好的答案，但需要改进。）

使用“TABLE”类型非常容易。让我们假设您的表名为Students，并且它具有列名。

declare @rowsCount INT
declare @i INT = 1
declare @names varchar(max) = ''

DECLARE @MyTable TABLE
(
  Id int identity,
  Name varchar(500)
)
insert into @MyTable select name from Students
set @rowsCount = (select COUNT(Id) from @MyTable)

while @i < @rowsCount
begin
 set @names = @names + ', ' + (select name from @MyTable where Id = @i)
 set @i = @i + 1
end
select @names

此示例使用SQL Server 2008 R2进行了测试。

Oracle有两种方法：

    create table name
    (first_name varchar2(30));

    insert into name values ('Peter');
    insert into name values ('Paul');
    insert into name values ('Mary');

解决方案是1：

    select substr(max(sys_connect_by_path (first_name, ',')),2) from (select rownum r, first_name from name ) n start with r=1 connect by prior r+1=r
    o/p=> Peter,Paul,Mary

解决方案是2:

    select  rtrim(xmlagg (xmlelement (e, first_name || ',')).extract ('//text()'), ',') first_name from name
    o/p=> Peter,Paul,Mary

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