in reply to How to escape parens for grep

The thing that starts with "tethereal " is a string (not a "command"). I think your post is a little confusing, because the string contains "6 or 17", but you say you want to assign "6,7", to a variable called $protocol.

Anyway, the way to set up a regex to capture the pieces you want involves ignoring or escaping the regex-special characters (the parens in this case) -- something like this would do: 

$_ = "tethereal (proto 6 or 17) and src host suntest1 and dst host sun
+test2 -w /tmp/dumpfile";

( /\(proto ([^)]+)\) and src host (\S+) and dst host (\S+) -w (\S+)/ )
    and ( $protocol, $srchost, $dsthost, $dir ) = ($1,$2,$3,$4);

$protocol =~ s/ or /,/g;

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(that is, if you want the value of $protocol to be a string consisting of comma-separated numbers, as opposed to something else)

Note the backslashes in front of the open and close parens that are intended to match actual parens in the string.

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Re^2: How to escape parens for grep
by swaroop (Beadle) on Aug 15, 2005 at 01:44 UTC
    Hi graff,

    Thanks for the reply. But sometimes the filters does't exists. I mean we may run with out specifying the proto / srchost / dsthost /dir.

    So, we need some generic solution to extract this command.

    Thanks again.
    - Swaroop
[reply]
      In that case you can split up the regex into multiple regex's and see if each one matches... 
      my $s = "tethereal (proto 6 or 17) and src host suntest1 and dst host 
+suntest2 -w /tmp/dumpfile";
my %info;
$info{srchost} = $1 if $s =~ /\bsrc host (\S+)/;
$info{dsthost} = $1 if $s =~ /\bdst host (\S+)/;
$info{dir} = $1 if $s =~ /\b-w (\S+)/;
if( $s =~ /\(proto (.*?)\)/ ){   # this one needs extra processing
  my $prot = $1;                          # start with the proto strin
+g: "6 or 17"
  my @prot = $prot =~ /(\d+)/g;           # pull out the integers:    
+   (6, 17)
  $info{protocol = join ',', @prot;       # create the desired string:
+   '6,17'
}
# if necessary, validate %info here to make sure you can continue...

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