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)

使用COALESCE-从这里了解更多信息

例如：

102103104

然后在SQL Server中编写以下代码，

Declare @Numbers AS Nvarchar(MAX) -- It must not be MAX if you have few numbers
SELECT  @Numbers = COALESCE(@Numbers + ',', '') + Number
FROM   TableName where Number IS NOT NULL

SELECT @Numbers

输出将是：

102,103,104

SQL Server 2005或更高版本

CREATE TABLE dbo.Students
(
    StudentId INT
    , Name VARCHAR(50)
    , CONSTRAINT PK_Students PRIMARY KEY (StudentId)
);

CREATE TABLE dbo.Subjects
(
    SubjectId INT
    , Name VARCHAR(50)
    , CONSTRAINT PK_Subjects PRIMARY KEY (SubjectId)
);

CREATE TABLE dbo.Schedules
(
    StudentId INT
    , SubjectId INT
    , CONSTRAINT PK__Schedule PRIMARY KEY (StudentId, SubjectId)
    , CONSTRAINT FK_Schedule_Students FOREIGN KEY (StudentId) REFERENCES dbo.Students (StudentId)
    , CONSTRAINT FK_Schedule_Subjects FOREIGN KEY (SubjectId) REFERENCES dbo.Subjects (SubjectId)
);

INSERT dbo.Students (StudentId, Name) VALUES
    (1, 'Mary')
    , (2, 'John')
    , (3, 'Sam')
    , (4, 'Alaina')
    , (5, 'Edward')
;

INSERT dbo.Subjects (SubjectId, Name) VALUES
    (1, 'Physics')
    , (2, 'Geography')
    , (3, 'French')
    , (4, 'Gymnastics')
;

INSERT dbo.Schedules (StudentId, SubjectId) VALUES
    (1, 1)        --Mary, Physics
    , (2, 1)    --John, Physics
    , (3, 1)    --Sam, Physics
    , (4, 2)    --Alaina, Geography
    , (5, 2)    --Edward, Geography
;

SELECT
    sub.SubjectId
    , sub.Name AS [SubjectName]
    , ISNULL( x.Students, '') AS Students
FROM
    dbo.Subjects sub
    OUTER APPLY
    (
        SELECT
            CASE ROW_NUMBER() OVER (ORDER BY stu.Name) WHEN 1 THEN '' ELSE ', ' END
            + stu.Name
        FROM
            dbo.Students stu
            INNER JOIN dbo.Schedules sch
                ON stu.StudentId = sch.StudentId
        WHERE
            sch.SubjectId = sub.SubjectId
        ORDER BY
            stu.Name
        FOR XML PATH('')
    ) x (Students)
;

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]

对于Oracle DB，请参阅以下问题：如何在Oracle中将多行连接成一行而不创建存储过程？

最好的答案似乎是@Emmanuel，使用了内置的LISTAGG（）函数，该函数在Oracle 11g Release 2和更高版本中提供。

SELECT question_id,
   LISTAGG(element_id, ',') WITHIN GROUP (ORDER BY element_id)
FROM YOUR_TABLE;
GROUP BY question_id

正如@user762952所指出的，根据Oracle的文档http://www.oracle-base.com/articles/misc/string-aggregation-techniques.php，WM_CONCAT（）函数也是一个选项。它看起来很稳定，但Oracle明确建议不要将它用于任何应用程序SQL，因此使用时风险自负。

除此之外，您还必须编写自己的函数；上面的Oracle文档提供了如何做到这一点的指南。

在MySQL中，有一个函数GROUP_CONCATT（），它允许您连接多行的值。例子：

SELECT 1 AS a, GROUP_CONCAT(name ORDER BY name ASC SEPARATOR ', ') AS people 
FROM users 
WHERE id IN (1,2,3) 
GROUP BY a