Question

I need to find a numeric value + quotes within a string (example 5″). Once found, I need to replace the quote with “pound”. Unfortunately, the string is complex:
Sample string

Work and run with Joe "The King" Mel using a 3" vest with "Dog-Bone" weights.

I have tried

SELECT REPLACE(REPLACE('Work and run with Joe "The King" Mel using a 3" vest with "Dog-Bone" weights.', '%[^0-9]"%', 'pound'), '"', 'pound')

But it replaces all the quotes with ‘pound’.

Answer 1

This will find the first occurrence of a number followed by a " and replace " with pound.

declare @s varchar(100) 
set @s = 'Work and run with Joe "The King" Mel using a 3" vest with "Dog-Bone" weights.'
select stuff(@s, patindex('%[0-9]"%', @s)+1, 1, ' pound')

STUFF

PATINDEX

If you have more than one you can put it in a while loop.

while patindex('%[0-9]"%', @s) > 0
begin
  set @s = stuff(@s, patindex('%[0-9]"%', @s)+1, 1, ' pound')
end