SPLIT string functions in SQL

I. Use a temporary table as an array
Copy codeThe Code is as follows:
Create function f_split (@ c varchar (2000), @ 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 affected rows is 4)


2. Split the string by the specified symbol and return the number of elements after the split. The method is very simple, that is, to check the number of separators in the string, and then add one, which is the required result.
Copy codeThe Code is as follows:
CREATE function Get_StrArrayLength
(
@ Str varchar (1024), -- string to be split
@ Split varchar (10) -- delimiter number
)
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

Call example: select dbo. Get_StrArrayLength ('78, 1, 2, 3 ',',')
Return Value: 4

3. Split the string by the specified symbol. Return the nth element of the specified index after the split, which is as convenient as an array.
Copy codeThe Code is as follows:
CREATE function Get_StrArrayStrOfIndex
(
@ Str varchar (1024), -- string to be split
@ Split varchar (10), -- Separator
@ Index int -- obtains the nth element.
)
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 cases: 1. There is no Separator in the string. 2. There is a separator in the string. After jumping out of the while LOOP, @ location is 0, by default, there is a separator behind the string.

Return substring (@ str, @ start, @ location-@ start)
End

Call example: select dbo. Get_StrArrayStrOfIndex ('8, 9, 4', ',', 2)
Return Value: 9

3. Combine the above two functions to traverse the elements in the string like an array
Copy codeThe Code is 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