Regex in SQL Server Replace function - sql-server

Question

I have a variable with random text, let's say
DECLARE #sNumberFormat NVARCHAR(200) = 'rand{text.here,{999}also-Random9He8re'
I want to replace each 9 in {999} by [0-9]. So in this example I would like to get
'rand{text.here,[0-9][0-9][0-9]also-Random9He8re'
Problem is I never know how many 9 will be placed in brackets, so there can be {99} {9999} ..and go on. I also need to validate if there is any invalid character (not 9) then nothing should be replaced.
I have tried some combinations of REPLACE and PATINDEX functions, but I could not achieve that.

Answer 1

Sans robust regex support, SQL Server's native functions do not give much help here. One approach, a bit hackish, would be to separate the input string into three components:
rand{text.here,
{999}
also-Random9He8re
Next, replace the 9 in the middle target substring with #, or some other character which you don't expect to appear anywhere else in your input string:
rand{text.here,
{###}
also-Random9He8re
Finally, replace the # in the middle substring with [0-9] and then concatenate together to get the final result:
DECLARE #val NVARCHAR(200) = 'rand{text.here,{999}also-Random9He8re'
SELECT REPLACE(
SUBSTRING(#val, 1, CHARINDEX('{9', #val) - 1) +
REPLACE(SUBSTRING(#val,
CHARINDEX('{9', #val) + 1,
CHARINDEX('9}', #val) - CHARINDEX('{9', #val)), '9', '#') +
SUBSTRING(#val, CHARINDEX('9}', #val) + 2, LEN(#val) - CHARINDEX('9}', #val)),
'#', '[0-9]');

Answer 2

So the lazy dev in me suggests this:
SELECT Replace(
Replace(
Replace(
Replace(#input, '{9999}', '[0-9][0-9][0-9][0-9]')
, '{999}', '[0-9][0-9][0-9]')
, '{99}', '[0-9][0-9]')
, '{9}', '[0-9]') AS result
;
You can keep extending as long as you like to perform your (one off?) replacements.
Quick. Simple. Extensible. Hacky.
Sometimes lazy is good enough.

Answer 3

This could be done with CTE series. It works with an arbitrary number of "9" values in square brackets.
Declare #str varchar(max) = 'rand{text.here,{999}also-Random9He8re';
With A As
(Select 1 As Pos
Union All
Select Pos+1 As Pos From A Where Pos < LEN(#str)
),
B As (
Select STRING_AGG(Case When Chr Like '[{9}]' Then Chr Else ' ' End, '') As Chr
From A Cross Apply (Select SUBSTRING(#str,A.Pos,1 )) As T(chr)
),
C As (
Select [value] As pattern,
REPLACE(REPLACE(REPLACE([value], '9', '[0-9]'),'{',''),'}','') As replacement,
ROW_NUMBER() Over (ORDER BY (SELECT NULL)) As Num,
COUNT(*) OVER (ORDER BY (SELECT NULL)) As Cnt
From B Cross Apply STRING_SPLIT(Chr,' ')
Where [value] Like '{%}' And [value] Like '%9%'
),
D As (
Select #str As Result, 1 As Num
Union All
select REPLACE(Result, C.pattern, C.replacement) As Res , D.Num+1 As Num
From D Inner Join C On (D.Num=C.Num)
Where D.Num<=C.Cnt)
Select Top 1 Result
From D
Order by Num Desc
A - Getting a list of character positions in text
B - Getting text with spaces instead of characters other than
'9','{','}'
C- Getting patterns and corresponding replacement values
D - Getting the result using REPLACEMENT function