问题
I'm using Perl to try and pull the IP address and port from a string.
The regex I'm trying to use is:
s/(sip:)(;.*)//
The strings are:
sip:255.255.255.255:8080;transport=TCP
sip:255.255.255.255:8080
Obviously my regex replace is not working. Can anyone point me to how I would need to write the regex? From those two strings I want to pull this:
255.255.255.255:8080
Meaning the regex needs to match anything that's not that string and replace it with nothing.
Note: IP address and port will be different every time.
回答1:
You can use the following:
/((\d{1,3}.){3}\d{1,3}:\d+)/
Explanation:
(\d{1,3}\.){3}
: between 1 and 3 consecutive digits followed by a period 3 times in a row
\d{1,3}:\d+
: between 1 and 3 consecutive digits followed by a colon followed by at least 1 digit
So in code it would look something like this:
my $s = 'sip:255.255.255.255:8080;transport=TCP';
my ($socket) = $s =~ /((\d{1,3}.){3}\d{1,3}:\d+)/;
回答2:
Here is a solution that uses the Regexp::Common
library to match the IP address.
use 5.022;
use Regexp::Common qw/net/;
while (<DATA>){
my @matches = ($_ =~ m/^sip:$RE{net}{IPv4}{-keep}:(\d+);?/);
my $ip = $matches[0].':'. $matches[5];
say $ip
}
__DATA__
sip:255.255.255.255:8080;transport=TCP
sip:255.255.255.255:8080
回答3:
It's not bulletproof, but this should work:
my $test = 'sip:255.255.255.255:8080;transport=TCP';
my ($ip) = $test =~ /(\d+\.\d+\.\d+\.\d+:\d+)/;
print "$ip\n";
This just extracts it, of course... you could always replace it this way:
$test = $ip;
回答4:
the regular expression tries to match your whole string which does not work since the wanted part is in between the parts you want to remove. so you either need 2 regular expressions to remove first the start and then the end or you need to use capture groups to get the part you want and replace with the captured part like this (not checked for correctness):
s/sip:(.*);.*/$1/
来源:https://stackoverflow.com/questions/34438566/regex-pulling-ip-and-port-from-string