正则表达式捕获IP地址（S，G）多播

Question

This regular expression works to capture IP addresses. I need one to capture this format:
(1.1.1.1,230.1.1.1)

How do I find a proper RegEx?

I would like to extract (S,G) as:

1.1.1.1 230.1.1.1

(...)
match = re.findall(r'^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$' , line)
(...)

Answer 1

You already have a pattern for one IP address. Now that you want to find a parenthesized pair of IP addresses, you could just repeat that pattern, putting , in between and \\( \\) . If you want to search multiple lines, you may want to add the multiline flag (?m) . To actually capture each address en bloc, you have to enclose it as one more group. This would make:

match = re.findall(r'(?m)^\((((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))'
                         +',(((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))\)$' , line)
for m in match:
    S = m[0]
    G = m[4]
    print S, G

That is of course ugly. We can improve it in a way by factoring out the repeated parts, eg:

I = '(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)'  # pattern for 1 to 255
IP = '(?:' +I+ '\.){3}' +I                      # pattern for IP address
SG = '(?m)^\((' +IP+ '),(' +IP+ ')\)$'          # pattern for (S,G)
match = re.findall(SG, line);
for S, G in match:
    print S, G

Here I also inserted ?: in groups which don't need to be retrieved, so that only the IP addresses remain in the match .