2

次の表から、Values列を複数の列に変換し、現在コンマで区切られている個々の値を入力するにはどうすればよいですか？変換前：

Name  Values 
----  ------
John  val,val2,val3 
Peter val5,val7,val9,val14 
Lesli val8,val34,val36,val65,val71,val 
Amy   val3,val5,val99

変換の結果は次のようになります。

Name  Col1  Col2  Col3  Col4  Col5  Col6 
----  ----  ----  ----  ----  ----  ----
John  val   val2  val3 
Peter val5  val7  val9  val14 
Lesli val8  val34 val36 val65 val71 val 
Amy   val3  val5  val99
4

2 に答える 2

3

これは、再帰的な cte を使用して「数値のテーブル」を生成するソリューションです( Itzik Ben-Ganの厚意による)。これは、文字列の分割や PIVOT を含むあらゆる種類の問題に役立ちます。SQL Server 2005 以降。完全なテーブルの作成、挿入、および選択スクリプトが含まれています。

CREATE TABLE dbo.Table1 
(
    Name        VARCHAR(30),
    [Values]    VARCHAR(128)
)
GO

INSERT INTO dbo.Table1 VALUES ('John', 'val,val2,val3')
INSERT INTO dbo.Table1 VALUES ('Peter', 'val5,val7,val9,val14')
INSERT INTO dbo.Table1 VALUES ('Lesli', 'val8,val34,val36,val65,val71,val')
INSERT INTO dbo.Table1 VALUES ('Amy', 'val3,val5,val99')
GO

SELECT * FROM dbo.Table1;
GO

WITH
L0 AS(SELECT 1 AS c UNION ALL SELECT 1),
L1 AS(SELECT 1 AS c FROM L0 AS A, L0 AS B),
L2 AS(SELECT 1 AS c FROM L1 AS A, L1 AS B),
L3 AS(SELECT 1 AS c FROM L2 AS A, L2 AS B),
Numbers AS(SELECT ROW_NUMBER() OVER(ORDER BY c) AS n FROM L3)
SELECT Name, [1] AS Column1, [2] AS Column2, [3] AS Column3, [4] AS Column4, [5] AS Column5, [6] AS Column6, [7] AS Column7
FROM
(SELECT Name,
        ROW_NUMBER() OVER (PARTITION BY Name ORDER BY nums.n) AS PositionInList,
        LTRIM(RTRIM(SUBSTRING(valueTable.[Values], nums.n, charindex(N',', valueTable.[Values] + N',', nums.n) - nums.n))) AS [Value]
 FROM   Numbers AS nums INNER JOIN dbo.Table1 AS valueTable ON nums.n <= CONVERT(int, LEN(valueTable.[Values])) AND SUBSTRING(N',' + valueTable.[Values], n, 1) = N',') AS SourceTable
PIVOT
(
MAX([VALUE]) FOR PositionInList IN ([1], [2], [3], [4], [5], [6], [7])
) AS Table2
GO

--DROP TABLE dbo.Table1

この出力を変換するもの

Name   Values
John   val,val2,val3
Peter  val5,val7,val9,val14
Lesli  val8,val34,val36,val65,val71,val
Amy    val3,val5,val99

に

Name  Column1 Column2 Column3 Column4 Column5 Column6 Column7
Amy   val3    val5    val99   NULL    NULL    NULL    NULL
John  val     val2    val3    NULL    NULL    NULL    NULL
Lesli val8    val34   val36   val65   val71   val     NULL
Peter val5    val7    val9    val14   NULL    NULL    NULL
于 2010-11-17T03:21:27.067 に答える
3

まず、使用しているデータベース製品とバージョンは何ですか? SQL Server 2005 以降を使用している場合は、Split ユーザー定義関数を次のように記述できます。

CREATE FUNCTION [dbo].[Split]
(   
    @DelimitedList nvarchar(max)
    , @Delimiter varchar(2) = ','
)
RETURNS TABLE 
AS
RETURN 
    (
    With CorrectedList As
        (
        Select Case When Left(@DelimitedList, DataLength(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End
            + @DelimitedList
            + Case When Right(@DelimitedList, DataLength(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End
            As List
            , DataLength(@Delimiter) As DelimiterLen
        )
        , Numbers As 
        (
        Select TOP (Coalesce(Len(@DelimitedList),1)) Row_Number() Over ( Order By c1.object_id ) As Value
        From sys.objects As c1
            Cross Join sys.columns As c2
        )
    Select CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen As Position
        , Substring (
                    CL.List
                    , CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen     
                    , CharIndex(@Delimiter, CL.list, N.Value + 1)                           
                        - ( CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen ) 
                    ) As Value
    From CorrectedList As CL
        Cross Join Numbers As N
    Where N.Value < Len(CL.List)
        And Substring(CL.List, N.Value, CL.DelimiterLen) = @Delimiter
    )

次に、次のようなものを使用して、必要な値を分割できます。

Select Name, Values
From Table1 As T1
Where Exists    (
                Select 1
                From Table2 As T2
                    Cross Apply dbo.Split (T1.Values, ',') As T1Values
                    Cross Apply dbo.Split (T2.Values, ',') As T2Values
                Where T2.Values.Value = T1Values.Value
                    And T1.Name = T2.Name
                )
于 2010-11-17T02:19:40.890 に答える