T-SQL - convertir une colonne délimitée par des virgules en plusieurs colonnes

Question

Dans le tableau ci-dessous, comment puis-je convertir la colonne Values en plusieurs colonnes, avec des valeurs individuelles séparées par des virgules? Avant la conversion:

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

Le résultat de la conversion devrait ressembler à:

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 

Answer 1

Tout d'abord, ce produit de base de données et la version que vous utilisez? Si vous utilisez SQL Server 2005 et plus tard, vous pouvez écrire une fonction définie par l'utilisateur de Split comme ceci:

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 
    ) 

Vous pouvez ensuite diviser les valeurs que vous voulez utiliser quelque chose semblable à:

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 
       ) 

Answer 2

Voici une solution qui utilise un cte récursif pour générer un "table of numbers" (courtoisie de Itzik Ben-Gan), qui est useful pour toutes sortes de problèmes, y compris le fractionnement de chaîne, et PIVOT. SQL Server 2005 et versions ultérieures. Table complète créer, insérer et sélectionner le script inclus.

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 

qui convertit cette sortie

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

à

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