One, using a temporary table as an array
Copy Code code as follows:
Create function F_split (@c varchar), @split varchar (2))
Returns @t table (col varchar (20))
As
Begin
while (CHARINDEX (@split, @c) <>0)
Begin
Insert @t (COL) VALUES (substring (@c,1,charindex (@split, @c)-1)
Set @c = Stuff (@c,1,charindex (@split, @c), "")
End
Insert @t (COL) VALUES (@c)
Return
End
Go
SELECT * from Dbo.f_split (' dfkd,dfdkdf,dfdkf,dffjk ', ', ')
Drop function F_split
Col
--------------------
Dfkd
Dfdkdf
Dfdkf
Dffjk
(The number of rows affected is 4 rows)
Second, according to the specified symbol segmentation string, return the number of elements after the split, the method is very simple, is to see how many delimiters in the string, and then add one, is the result of the request.
Copy Code code as follows:
CREATE function Get_strarraylength
(
@str varchar (1024),--string to be split
@split varchar (10)--Separator symbol
)
returns int
As
Begin
DECLARE @location int
DECLARE @start int
DECLARE @length int
Set @str =ltrim (RTrim (@str))
Set @location =charindex (@split, @str)
Set @length =1
While @location <>0
Begin
Set @start = @location +1
Set @location =charindex (@split, @str, @start)
Set @length = @length +1
End
Return @length
End
Invocation Example: SELECT dbo. Get_strarraylength (' 78,1,2,3 ', ', ', ')
return value: 4
Dividing a string by a specified symbol, returning the first element of the specified index after the partition, as convenient as an array
Copy Code code as follows:
CREATE function Get_strarraystrofindex
(
@str varchar (1024),--string to be split
@split varchar (10),--Separator symbol
@index INT--Take the first few elements
)
Returns varchar (1024)
As
Begin
DECLARE @location int
DECLARE @start int
DECLARE @next int
DECLARE @seed int
Set @str =ltrim (RTrim (@str))
Set @start =1
Set @next =1
Set @seed =len (@split)
Set @location =charindex (@split, @str)
While @location <>0 and @index > @next
Begin
Set @start = @location + @seed
Set @location =charindex (@split, @str, @start)
Set @next = @next +1
End
If @location =0 select @location =len (@str) +1
There are two conditions: 1, the string does not exist separator symbol 2, there is a delimiter in the string, jumping out of the while loop, the @location is 0, the default is a string with a delimiter behind.
return substring (@str, @start, @location-@start)
End
Invocation Example: SELECT dbo. Get_strarraystrofindex (' 8,9,4 ', ', ', ', 2)
Return Value: 9
Three, combining the top two functions, like arrays to traverse the elements in the string
Copy Code code as follows:
DECLARE @str varchar (50)
Set @str = ' 1,2,3,4,5 '
DECLARE @next int
Set @next =1
While @next <=dbo. Get_strarraylength (@str, ', ')
Begin
Print dbo. Get_strarraystrofindex (@str, ', ', @next)
Set @next = @next +1
End
Call Result:
1
2
3
4
5
