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“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。

我在家里无法访问SQL Server，所以我猜测这里的语法，但大致上是这样的：

DECLARE @names VARCHAR(500)

SELECT @names = @names + ' ' + Name
FROM Names

以下是实现这一目标的完整解决方案：

-- Table Creation
CREATE TABLE Tbl
( CustomerCode    VARCHAR(50)
, CustomerName    VARCHAR(50)
, Type VARCHAR(50)
,Items    VARCHAR(50)
)

insert into Tbl
SELECT 'C0001','Thomas','BREAKFAST','Milk'
union SELECT 'C0001','Thomas','BREAKFAST','Bread'
union SELECT 'C0001','Thomas','BREAKFAST','Egg'
union SELECT 'C0001','Thomas','LUNCH','Rice'
union SELECT 'C0001','Thomas','LUNCH','Fish Curry'
union SELECT 'C0001','Thomas','LUNCH','Lessy'
union SELECT 'C0002','JOSEPH','BREAKFAST','Bread'
union SELECT 'C0002','JOSEPH','BREAKFAST','Jam'
union SELECT 'C0002','JOSEPH','BREAKFAST','Tea'
union SELECT 'C0002','JOSEPH','Supper','Tea'
union SELECT 'C0002','JOSEPH','Brunch','Roti'

-- function creation
GO
CREATE  FUNCTION [dbo].[fn_GetItemsByType]
(   
    @CustomerCode VARCHAR(50)
    ,@Type VARCHAR(50)
)
RETURNS @ItemType TABLE  ( Items VARCHAR(5000) )
AS
BEGIN

        INSERT INTO @ItemType(Items)
    SELECT  STUFF((SELECT distinct ',' + [Items]
         FROM Tbl 
         WHERE CustomerCode = @CustomerCode
            AND Type=@Type
            FOR XML PATH(''))
        ,1,1,'') as  Items



    RETURN 
END

GO

-- fianl Query
DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(Type) 
                    from Tbl
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT CustomerCode,CustomerName,' + @cols + '
             from 
             (
                select  
                    distinct CustomerCode
                    ,CustomerName
                    ,Type
                    ,F.Items
                    FROM Tbl T
                    CROSS APPLY [fn_GetItemsByType] (T.CustomerCode,T.Type) F
            ) x
            pivot 
            (
                max(Items)
                for Type in (' + @cols + ')
            ) p '

execute(@query) 

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]

SELECT STUFF((SELECT ', ' + name FROM [table] FOR XML PATH('')), 1, 2, '')

下面是一个示例：

DECLARE @t TABLE (name VARCHAR(10))
INSERT INTO @t VALUES ('Peter'), ('Paul'), ('Mary')
SELECT STUFF((SELECT ', ' + name FROM @t FOR XML PATH('')), 1, 2, '')
--Peter, Paul, Mary