Regex to extract IPs between () not working

[mrapple2020]

Hello, can you please help me retrieve IP addresses between (). Like this:
(1.1.1.1,230.1.1.1)

Notice the below is not working since it is capturing text like this: match=re.findall(r'\(([0-9].+)\),string) if match: match = my_s_g_ip (...)

It would also capture:
(source,group)

[Larz60+]

show runable snippet

[mrapple2020]

To be more precise, I am puzzled why the below doesn't match (50.50.50.22,230.207.200.20). Any ideas?

s = """ rate 230.207.200.1', '', ' (50.50.50.11,230.207.200.1)', ' Incoming rate : 1013 / 8127181 (0/5/CPU0)', ' Outgoing rate : ', ' Node : 5/6/CPU0 : 933 / 15828597. >> it shows double as O-list is on 2 PSE’s.', ' Node : 0/6/CPU0 : 1032 / 7966755 ', ' Node : 0/9/CPU0 : 1035 / 15828597. >> it shows double as O-list is on 2 PSE’s.', ' Node : 0/1/CPU0 : 998 / 7966755 ', ' Node : 5/6/CPU0 : 999 / 15828597. >> it shows double as O-list is on 2 PSE’s.', ' Node : 0/6/CPU0 : 899 / 7966755 ', ' Node : 0/9/CPU0 : 1012 / 15828597. >> it shows double as O-list is on 2 PSE’s.', ' Node : 0/1/CPU0 : 999 / 7966755 ', ' [b](50.50.50.22,230.207.200.22)[/b]', ' Incoming rate : 2222 / 8127181 (0/5/CPU0)', ' Outgoing rate : ', ' Node : 1/3/CPU0 : 933 / 15828597. >> it shows double as O-list is on 2 PSE’s.', ' Node : 1/2/CPU0 : 1032 / 7966755 ', '', ' ', ' ', ' '] """ >>> match = re.findall(r'([0-9].[0-9].[0-9].[0-9],[0-9].[0-9].[0-9].[0-9].[0-9])', s) >>> match [] >>> >>> match = re.findall(r'([0-9].[0-9].[0-9].[0-9],[0-9].[0-9].[0-9].[0-9].[0-9])', s) >>> match [] >>>

[metulburr]

. is a wildcard in regex to stand for every character

try this regex

\(\d{1,3}(\.\d{1,3}){3},\d{1,3}(\.\d{1,3}){3}\)

https://www.regextester.com/?fam=108886

[mrapple2020]

It improved. At least I get match. But any idea why I am getting only incomplete like below? I try on rubular.com and it appears to work. But when I apply it to my text it is different:

>>> >>> s " #show mfib route rate 230.207.200.1', '', ' (50.50.50.11,230.207.200.1)', " >>>for item in s: match = re.findall(r'\(\d{1,3}(\.\d{1,3}){3},\d{1,3}(\.\d{1,3}){3}\)', item) if match: print(match) ... [('.11', '.1')] [('.11', '.1')] [('.11', '.1')] [('.11', '.1')] [('.11', '.1')] [('.11', '.1')] [('.11', '.1')] [('.11', '.1')]

[DeaD_EyE]

Are you iterating over a string?
Check, if item is really a whole line.

Then try this regex:

ip_block = r'((?:\d{1,3}\.){3}\d{1,3})' # ip is a group ip_line = rf'\({ip_block},{ip_block}\)' # parenthesis around ip1,ip2

The second one is a format string. It takes the ip_block twice spitted by comma and surrounded by parenthesis.
Finally the resulting regex:

Output:
\(((?:\d{1,3}\.){3}\d{1,3}),((?:\d{1,3}\.){3}\d{1,3})\)

Then iterate line by line over the string or file-object.
Use re.match(the_regex, line).
If re.match: -> then you have match.group(1) for the left ip and match.group(2) for the right ip.