Remove characters before and after string SQL - sql

Question

I would like to remove all characters before and after a string in the select statement.
In the example below I would like to remove everything before and including /Supply> and after and including >/
Note the remaining part will be a fixed number of characters.
Any help would be much appreciated
Eg.
abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd
Would become:
hhfhjgglldppprrr

Answer 1

If your input always has exactly two instances of ">" you could use PARSENAME.
declare #SomeValue varchar(100) = 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
select PARSENAME(replace(#SomeValue, '>', '.'), 2)
This will not work correctly if your data also contains any periods (.). We can deal with that if needed with a couple of replace statements. Still very simple and easy to maintain with the same caveat of exactly 2 >.
declare #SomeOtherValue varchar(100) = 'abs/Supply>hhfhjgg.lldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
select replace(PARSENAME(replace(replace(#SomeOtherValue, '.', '~!##'), '>', '.'), 2), '~!##', '.')

Answer 2

You can use PATINDEX() to identify the position of the patterns you are looking for (/Supply> and >/) then remove them based on the length of the string:
SELECT LEFT(RIGHT(col,LEN(col) - PATINDEX('%/Supply>%',col) -7), PATINDEX('%>/%', RIGHT(col,LEN(col) - PATINDEX('%Supply>%',col) -7))-1)
Simply replace col in the above with your column name.
Example below with test string abs/Supply>keep>/remove
First remove everything before and including /Supply>:
SELECT RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7)
This will give keep>/remove
Then remove everything after and including >/:
SELECT LEFT('keep>/remove',PATINDEX('%>/%','keep>/remove') - 1)
This will give keep, the part of the string you want.
Here is the combined version, same as above, just includes the test string instead of col so you can run it easily:
SELECT LEFT(RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7), PATINDEX('%>/%', RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7))-1)
This will give keep. You can also replace the string above with the one in your question, I just used a different test string because it is shorter and makes the code more readable.

Answer 3

try this:
DECLARE #inputStr VARCHAR(max)= 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
DECLARE #startString VARCHAR(100)='/Supply>'
DECLARE #EndString VARCHAR(100)='>/'
DECLARE #LenStartString INT = LEN(#startString)
DECLARe #TempInputString VARCHAR(max)='';
DECLARE #StartIndex INT
DECLARE #EndIndex INT
SELECT #StartIndex = CHARINDEX(#startString,#inputStr)+#LenStartString
SELECT #TempInputString = STUFF(#inputStr, 1, #StartIndex, '')
SELECT SUBSTRING(#TempInputString,0,CHARINDEX(#EndString,#TempInputString))
In Single Line
DECLARE #inputStr VARCHAR(max)= 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
DECLARE #startString VARCHAR(100)='/Supply>'
DECLARE #EndString VARCHAR(100)='>/'
SELECT SUBSTRING(STUFF(#inputStr, 1, CHARINDEX(#startString,#inputStr)+LEN(#startString), ''),0,CHARINDEX(#EndString,STUFF(#inputStr, 1,CHARINDEX(#startString,#inputStr)+LEN(#startString), '')))