How to groups() in regex with whitespace

Asked Sep 18 '19 at 18:30

Active Sep 18 '19 at 18:35

Viewed 15 times

I have a regex that i'm trying to match the phone number. However, 1 of the scenario of phone numbers have blanks/whites paces in the front and it's causing groups() error because of NoneType

How do I avoid this?

s = ' 1234567890'

pattern = re.compile(r'''^(?:\s*\(?)(\d{3})\)*[-\. ]*(\d{3})[-. ]?(\d{4})\s+$''',re.VERBOSE)
match = pattern.match(s)

match.groups()

AttributeError: 'NoneType' object has no attribute 'groups'

asked Sep 18 '19 at 18:30

user11666514

What about `pattern.match(s.strip())`? – Dani Mesejo Sep 18 '19 at 18:31
The string does not match. What are the requirements? Why `\s+` at the end? It requires 1+ whitespaces, but your input has no whitespace at the end. Use `\s*` – Wiktor Stribiżew Sep 18 '19 at 18:33
I'd want to break the phone number into groups.... s.strip() doesn't work – user11666514 Sep 18 '19 at 18:35
@WiktorStribiżew OMG That works....I didn't even realize the typo. Regex is tricky! – user11666514 Sep 18 '19 at 18:35
You may read about the difference at the [linked post](https://stackoverflow.com/questions/10763820/difference-between-and-regex). – Wiktor Stribiżew Sep 18 '19 at 18:36

How to groups() in regex with whitespace

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