考虑一个包含名称的数据库表,其中有三行:

Peter
Paul
Mary

有没有一种简单的方法可以把它变成彼得、保罗、玛丽的一串?


当前回答

我并没有做过任何关于性能的分析,因为我的列表中只有不到10项,但在查看了30多个答案后,我感到很惊讶,我仍然对已经给出的类似答案进行了修改,类似于对单个组列表使用COALESCE,甚至不必设置我的变量(无论如何默认为NULL),并且它假设我的源数据表中的所有条目都是非空的:

DECLARE @MyList VARCHAR(1000), @Delimiter CHAR(2) = ', '
SELECT @MyList = CASE WHEN @MyList > '' THEN @MyList + @Delimiter ELSE '' END + FieldToConcatenate FROM MyData

我确信COALENCE内部使用了相同的想法。让我们希望微软不会在我身上改变这一点。

其他回答

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)
;

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) 
   declare @phone varchar(max)='' 
   select @phone=@phone + mobileno +',' from  members
   select @phone

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

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)