How to match a regex expression only if a word is present before or after

Question

I’m really struggling with some regex. I’ve had a good look at similar questions and I can’t work out why it’s not working!

I’m trying to match the string ‘ok’ when it is preceded by 4 digits ((?<=d{4}s)ok) but only when the word ‘devo’ appears anywhere before or anywhere after in the string

text_to_search="devo XXXXXXXXXX 9999 ok ferial blabla" 
pattern = re.compile(r"(?<=devo)((?<=d{4}s)ok)|((?<=d{4}s)ok)(?=devo)")
matches = pattern.finditer (text_to_search)
[print (x) for x in matches]

This does not return any matches… if i try with:

text_to_search=" XXXXXXXXXX 9999 ok ferial blabla devo"

it does not work either.

Just for clarity, the two examples above should match, but if i take another example like :

text_to_search=" XXXXXXXXXX 9999 ok ferial blabla"

then this should not produce a match since ‘devo’ is not present..

Thank you in advance for your help

Asked By: tezzaaa

||

Source

Answer 1

You have nothing for the anywhere before or after part. Try this

(?<=devo).*((?<=d{4}s)ok)|((?<=d{4}s)ok).*(?=devo)

But I think the lookarounds are overkill here. A much simpler

(devo.*d{4}sok)|(d{4}sok.*devo)

might work as well.

Answered By: Paul Pazderski

Answer 2

You could make the regex easier by just checking if devo is in the string first:

import re

text_to_search=" XXXXXXXXXX 9999 ok ferial blabla devo" 

ok = re.compile(r'(?<=d{4}s)ok')

if 'devo' in text_to_search:
    matches = [m.string for m in ok.finditer(text_to_search)]
else:
    print('no match')
    matches = []

Answered By: C.Nivs

How to match a regex expression only if a word is present before or after

Question:

Answers: