Failed to create the right pattern to extract the desired portion from a string using regex

Question

What would be the right pattern if I applied regex to the following string:

item = "JavaScript:muestra('01','043','071','01','600%2D2023%2DSUNARP%2DTR')"

and the result I wish to get is:

0104307101600-2023-SUNARP-TR

I've tried with:

import re

item = "JavaScript:muestra('01','043','071','01','600%2D2023%2DSUNARP%2DTR')"
content = re.findall(r"(\('.+?'\))",item)[0].replace("'","").replace(",","")
print(content)

Answer 1

Try:

>>> import re
>>> item = "JavaScript:muestra('01','043','071','01','600%2D2023%2DSUNARP%2DTR')"
>>> chunks = re.findall(r"'([^']+)'", item)
>>> chunks
['01', '043', '071', '01', '600%2D2023%2DSUNARP%2DTR']
>>> chunks[-1]
'600%2D2023%2DSUNARP%2DTR'

If this feels too "loose", you can first grab the parenthesized area with something like:

paren = re.search(r"\(([^)]+)\)", item)

or, more precisely:

paren = re.search(r"JavaScript:muestra\(([^)]+)\)", item)

then run the '-based match in the top snippet.

Using re.split() can also work:

[x for x in re.split(r"'(?:,')?", paren[1]) if x]

Another approach is ast.literal_eval to convert the parenthesized substring into a tuple:

>>> import ast
>>> paren = re.search(r"\([^)]+\)", item)
>>> ast.literal_eval(paren[0])
('01', '043', '071', '01', '600%2D2023%2DSUNARP%2DTR')
>>> tup = ast.literal_eval(paren[0])
>>> tup[-1]
'600%2D2023%2DSUNARP%2DTR'

Boiling this down to a one-liner without regex:

>>> ast.literal_eval(item.replace("JavaScript:muestra", ""))[-1]
'600%2D2023%2DSUNARP%2DTR'