Clash Royale CLAN TAG#URR8PPP

What is the RE to match the list?

I want to know how to construct the regular express to extract the list.

Here is my string:

audit = "## audit_filter = ['hostname.*','service.*'] ##"


Here is my expression:

AUDIT_FILTER_RE = r'([.*])'


And here is my search statement:

audit_filter = re.search(AUDIT_FILTER_RE, audit).group(1)


I want to extract everything inside the square brackets including the brackets. '[...]'

Expected Output:

['hostname.*','service.*']


print(re.findall(r"[(.*?)]", audit))

([[^]]+])

2 Answers
2

import re
audit = "## audit_filter = ['hostname.*','service.*'] ##"
print eval(re.findall(r"[.*]", audit)[0]) # ['hostname.*', 'service.*']


findall returns a list of string matches. In your case, there should only be one, so I'm retrieving the string at index 0, which is a string representation of a list. Then, I use eval(...) to convert that string representation of a list to an actual list. Just beware:

findall

eval(...)

eval()

Use r"[(.*?)]"

r"[(.*?)]"

Ex:

import re
audit = "## audit_filter = ['hostname.*'] ##"
print(re.findall(r"[(.*?)]", audit))


Output:

["'hostname.*'"]


"RESULT"

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