Możesz to zrobić za pomocą funkcji Split.Poniższa funkcja Podział opiera się na istnieniu tabeli numery, które dosłownie zawiera sekwencyjny wykaz numerów tak:
Create Table dbo.Numbers(Value int not null primary key clustered)
GO
With Nums As
(
Select ROW_NUMBER() OVER(Order By o.object_id) As Num
From sys.objects as o
cross join sys.objects as o2
)
Insert dbo.Numbers(Value)
Select Num
From Nums
Where Num Between 1 And 10000
GO
Create Function [dbo].[udf_Split] (@DelimitedList nvarchar(max), @Delimiter nvarchar(2) = ',')
Returns @SplitResults TABLE (Position int NOT NULL PRIMARY KEY, Value nvarchar(max))
AS
/*
PURPOSE: to split the @DelimitedList based on the @Delimter
DESIGN NOTES:
1. In general the contents of the next item is: NextDelimiterPosition - CurrentStartPosition
2. CurrentStartPosition =
CharIndex(@Delimiter, A.list, N.Value) = Current Delimiter position
+ Len(@Delimiter) + The number of delimiter characters
+ 1 + 1 since the text of the item starts after the delimiter
3. We need to calculate the delimiter length because the LEN function excludes trailing spaces. Thus
if a delimiter of ", " (a comma followed by a space) is used, the LEN function will return 1.
4. The DataLength function returns the number of bytes in the string. However, since we're using
an nvarchar for the delimiter, the number of bytes will double the number of characters.
*/
Begin
Declare @DelimiterLength int
Set @DelimiterLength = DataLength(@Delimiter)/2
If Left(@DelimitedList, @DelimiterLength) <> @Delimiter
Set @DelimitedList = @Delimiter + @DelimitedList
If Right(@DelimitedList, @DelimiterLength) <> @Delimiter
Set @DelimitedList = @DelimitedList + @Delimiter
Insert @SplitResults(Position, Value)
Select CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength
, Substring (
A.List
, CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength
, CharIndex(@Delimiter, A.list, N.Value + 1)
- (CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength)
)
From dbo.Numbers As N
Cross Join (Select @DelimitedList As list) As A
Where N.Value > 0
And N.Value < LEN(A.list)
And Substring(A.list, N.Value, @DelimiterLength) = @Delimiter
Order By N.Value
Return
End
wtedy może być w stanie uruchomić kwerendę jak tak gdzie strip prefiksy:
Select Table, Substring(S.Value, CharIndex(':', S.Value) + 1, Len(S.Value))
From Table
Cross Apply dbo.udf_Split(Table.ListColumn, '/') As S
To daje wartości, takich jak:
Category1
Category2
Category3
można wtedy użyć FOR XML PATH
połączyć je ponownie:
Select Table.PK
, Stuff( (
Select '/' + Substring(S.Value, CharIndex(':', S.Value) + 1, Len(S.Value))
From Table As Table1
Cross Apply dbo.udf_Split(Table.ListColumn, '/') As S1
Where Table1.PK = Table.PK
Order By S1.Position
For Xml Path('')
), 1, 1, '') As BreadCrumb
From Table
Co produktów i wersji bazy danych? – Thomas
odebrano za pomocą prostego wybierania instrukcji za pomocą CTE, poniżej – Oliver