Matching an IP address

ibanix has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
(z) Re: Regex redux by zigdon (Deacon) on Nov 19, 2002 at 18:41 UTC
Well, assuming you want to stick with this regex for ip's (which allows things like 999.012.4231.31245 - see Matching an IP address), you could just define it beforehand: `my $ip = qr/\d+\.\d+\.\d+\.\d+/; foreach $line (@connections) { if ($line =~ /^\s($ip):\d+ \s->\s($ip):\d+ \s->\s*($ip)/x) { $client_ip{$1}{$iteration}++; $vips{$2}{$iteration}++; $frontend{$3}{$iteration}++; } }` [download] -- Dan	[reply] [d/l]
Re: Regex redux by dws (Chancellor) on Nov 19, 2002 at 19:30 UTC
Is there a better way I can do the (\d+\.\d+\.\d+\.\d+)? What you have is fine, though you could take more advantage of /x to clean up the regex: `if ( $line =~ m{ ^\s* ( # begin client IP \d+\.\d+\.\d+\.\d+ ) :\d+ # client port (ignored) \s->\s ( # begin vips \d+\.\d+\.\d+\.\d+ ) :\d+ # vips port (ignored) \s->\s ( # begin frontend \d+\.\d+\.\d+\.\d+ ) }x ) { $client_ip{$1}{$iteration}++; $vips{$2}{$iteration}++; $frontend{$3}{$iteration}++; }` [download] If you know the number of spaces around "->", use it instead of \s* (e.g., "\s->\s" instead of "\s->\s"). If you've got a lot of data, you're probably not going to want to pull it all into @connections. That's gotta suck up RAM. Also, consider inverting the data structure you're collecting the counts in. If $iteration is relatively fixed (i.e., changes slowly, compared to the number of connections you're processing), you might save significant time by taking counts without considering $iteration, then sweep those counts into a larger data structure whenever $iteration changes. This is one to benchmark, since it could easily backfire depending on your data mix.	[reply] [d/l]
Re: Re: Regex redux by ibanix (Hermit) on Nov 19, 2002 at 20:55 UTC
The data from @connections is thankfully small. It's the endless upper bounds of $iteration that should suck up RAM in the long run. I haven't figured how I should deal with that yet. I've posted the full script (and questions) at http://www.perlmonks.org/index.pl?node_id=214252 Thanks! <-> In general, we find that those who disparage a given operating system, language, or philosophy have never had to use it in pratice. <->	[reply]
Re: Regex redux by shemp (Deacon) on Nov 19, 2002 at 18:47 UTC
not the absolute greatest reduction, but: `((?:\d+\.){3}\d+)` [download] The IP is still stored in $1, since the inner parentheses has the ?: directive so that value isnt captured, but the match to the outer parentheses goes in $1	[reply] [d/l]
Re: Regex redux by adrianh (Chancellor) on Nov 19, 2002 at 21:52 UTC
Is there a better way I can do the (\d+\.\d+\.\d+\.\d+)? As you can tell, I'm just trying to match for a numeric IP address. Take a look at Regexp::Common::net. Your code could be written as something like (untested code): `use Regexp::Common qw /net/; # match dotted decimal IP address # $1 = whole match, $2-5 = bytes my $IP = $RE{net}{IPv4}{-keep}; foreach my $line (@connections) { if ($line =~ /^\s$IP:\d+\s->\s$IP:\d+\s->\s*$IP/x) { $client_ip{$1}{$iteration}++; $vips{$6}{$iteration}++; $frontend{$11}{$iteration}++; } }` [download] Note that $IP will only match on legal decimal IP addresses (so 666.666.666.666 won't match). This may, or may not, be what you want ;-)	[reply] [d/l]
Re: Regex redux by petral (Curate) on Nov 20, 2002 at 02:36 UTC
`if ( @ips = $line =~ /[\d.]{7,15}/g ) { $client_ip{$ips[0]}{$iteration}++; . . . }` [download] If you need to check the validity of the line, you can wrap that in something like (using the regex from Mastering Reg Exps as cited here): `$numrx = qr/[01]?\d\d?\|2[0-4]\d\|25[0-5]/; $iprx = qr/($numrx\.){3}$numrx/; if ( $line =~ /^(\s($iprx):\d+\s->){2}\s$iprx/ ) { . . . }` [download] Or, of course combine them: `if ( $line =~ /^\s($iprx):\d+\s->\s($iprx):\d+\s->\s($iprx)/ ) {` p	[reply] [d/l] [select]


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