Extract values between square brackets ignoring single quotes

Question

I am trying the following:

s = "Text text text [123] ['text']"

This is my function:

def getFromSquareBrackets(s):
    m = re.findall(r"\[([A-Za-z0-9_']+)\]", s)
    return m

But I am obtaining:

['123', "'text'"]

I want to obtain:

['123', 'text'] 

How can I ignore the single quotes?

Answer 1

You can make ' optional using ? as

>>> re.findall(r"\['?([^'\]]+)'?\]", s)
['123', 'text']

---

• \['? Matches [ or ['.

• ([^'\]]+) Matches anything other than ' or ] and captures them.

• '?\] Matches ] or ']

Answer 2

Make ' optional and outside the capturing group

m = re.findall(r"\['?([A-Za-z0-9_]+)'?\]", s)