Python - re.findall returns unwanted result
findall, python, regex
Solution
The trivial solution:
>>> re.findall("(100%|[0-9][0-9]%|[0-9]%)","89%")
['89%']
More beautiful solution:
>>> re.findall("(100%|[0-9]{1,2}%)","89%")
['89%']
The prettiest solution:
>>> re.findall("(?:100|[0-9]{1,2})%","89%")
['89%']
Problem
``` re.findall("(100|[0-9][0-9]|[0-9])%", "89%") ``` This returns only result `[89]` and I need to return the whole 89%. Any ideas how to do it please?