Split a string and retain character position - sql-server

Question

I have a large character string (nvarchar(MAX)) that I am trying to split 2 and identify which part of the original string they are.
Example
String:
5;718;0;1071;1.23|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25
I would first need to split the strings based on the '|' character so something like this:
5;718;0;1071;1.23
0;750;0;997;1.25
0;750;0;997;1.25
0;750;0;997;1.25
0;750;0;997;1.25
0;750;0;997;1.25
I would then to split each of those based on the ';' character:
So 5;718;0;1071;1.23 would then split into:
5
718
0
1071
1.23
I know I could do a string_split on the '|' then another string_split on the ';' but that does not maintain an order or identify from which portion of the string the result is split from and I am unfortunately not quite able to get the results I am looking for when trying to use OPENJSON():
Based on the example above I would need a result that could identify that the 718 was from the first group and the 2nd item in said group.

Answer 1

Here is an option that will parse your string and maintain the sequence
Example
Declare #S varchar(max) = '5;718;0;1071;1.23|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25'
Select Seq1 = A.RetSeq
,Val1 = A.RetVal
,Seq2 = B.RetSeq
,Val2 = B.RetVal
From [dbo].[tvf-Str-Parse](#S,'|') A
Cross Apply [dbo].[tvf-Str-Parse](A.RetVal,';') B
Returns
The Function if Interested
CREATE FUNCTION [dbo].[tvf-Str-Parse] (#String varchar(max),#Delimiter varchar(10))
Returns Table
As
Return (
Select RetSeq = row_number() over (order by 1/0)
,RetVal = ltrim(rtrim(B.i.value('(./text())[1]', 'varchar(max)')))
From (Select x = Cast('<x>' + replace((Select replace(#String,#Delimiter,'§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A
Cross Apply x.nodes('x') AS B(i)
);
EDIT - Update for TABLE
Declare #YourTable table (ID int,SomeCol varchar(max))
Insert Into #YourTable values
(1,'5;718;0;1071;1.23|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25')
Select A.ID
,Seq1 = B.RetSeq
,Val1 = B.RetVal
,Seq2 = C.RetSeq
,Val2 = C.RetVal
From #YourTable A
Cross Apply [dbo].[tvf-Str-Parse](SomeCol,'|') B
Cross Apply [dbo].[tvf-Str-Parse](B.RetVal,';') C

Answer 2

here's my old school take:
declare #v nvarchar(max) = N'5;718;0;1071;1.23|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25'
select a.value('.', 'nvarchar(max)') [value], v2.col, rnInternal,ROW_NUMBER() over(partition by rnInternal order by rnInternal, Split.a) rnExternal
from
(
select cast('<M>' + REPLACE(v.[value], ';', '</M><M>') + '</M>' AS XML) as col, rnInternal
from
(
select
a.value('.', 'nvarchar(max)') [value],
ROW_NUMBER() over(order by Split.a) rnInternal
from
(select cast('<M>' + REPLACE(#v, '|', '</M><M>') + '</M>' AS XML) as col) as A
CROSS APPLY A.col.nodes ('/M') AS Split(a)
where
a.value('.', 'nvarchar(max)') <> ''
) v
) v2
CROSS APPLY v2.col.nodes ('/M') AS Split(a)
order by rnInternal, rnExternal

Answer 3

You are right about the STRING_SPLIT() (as is mentioned in the documentation the output rows might be in any order), but you may use a JSON-based approach to get the expected results.
You need to transform the input text into a valid nested JSON array (5;718;0;1071;1.23|0;750;0;997;1.25| into [[5,718,0,1071,1.23],[0,750,0,997,1.25]]) and parse this array with two OPENJSON() calls using a default schema. The result from the OPENJSON() call is a table with columns key, value and type and in case of JSON array the key column returns the index of the element in the specified array:
DECLARE #text nvarchar(max) = N'5;718;0;1071;1.23|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25|0;750;0;997;1.25'
SELECT
CONVERT(int, j1.[key]) + 1 AS id1,
CONVERT(int, j2.[key]) + 1 AS id2,
j2.[value]
FROM OPENJSON(CONCAT('[[', REPLACE(REPLACE(#text, '|', '],['), ';', ','), ']]')) j1
CROSS APPLY OPENJSON(j1.[value]) j2
ORDER BY CONVERT(int, j1.[key]), CONVERT(int, j2.[key])
Result:
id1 id2 value
1 1 5
1 2 718
1 3 0
1 4 1071
1 5 1.23
2 1 0
2 2 750
...
6 1 0
6 2 750
6 3 0
6 4 997
6 5 1.25
If you want to identify the id's of each group and subgroup, you need to add an appropriate WHERE clause:
SELECT
CONVERT(int, j1.[key]) + 1 AS id1,
CONVERT(int, j2.[key]) + 1 AS id2
FROM OPENJSON(CONCAT('[[', REPLACE(REPLACE(#text, '|', '],['), ';', ','), ']]')) j1
CROSS APPLY OPENJSON(j1.[value]) j2
WHERE j2.[value] = '718'
ORDER BY CONVERT(int, j1.[key]), CONVERT(int, j2.[key])
Result:
id1 id2
1 2