New to perl and regex

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

Hello, I'm new to Perl and regex and i cant seem to get this expression to work, I'm trying to match any amount of white space 0 or 1 times followed by any amount of words saved in a buffer(there can be any amount of space in between the words), after the words there has to be at least 1 or more white space and followed by a digit that ranges from -99 to 99 here's what I have wrote down.

print "Enter a string: ";
while(<>)
{
    $line = $_;
 last if ($_ =~ /^\s*[Q|q]{1,1}[U|u]{1,1}[I|i]{1,1}[T|t{1,1}\s*$/);
    next if ($_ =~ /^$/ or $_ =~ /^\s+/$);
    if($line =~ /\s*([a-zA-Z]+\s*)+\s*(-?\d\d?)\s*/)
    {
    print $_;
    }
}
[download]

this if($line =~ /\s*([a-zA-Z]+\s*)+\s*(-?\d\d?)\s*/) is where the problem comes in. Is there any notable problems here that I'm over looking?

Comment on New to perl and regex Select or Download Code

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Re: New to perl and regex by NetWallah (Canon) on Nov 19, 2012 at 02:17 UTC
Welcome to the Monastery ! You have other problems with unmatched "[" and the strange "$" as a regex modifier. I have fixed that - and modified your target selelection slightly by adding a trailing "$". `use strict; use warnings; print "Enter a string: "; while(<>) { my $line = $_; last if ($_ =~ /^\s[Q\|q]{1,1}[U\|u]{1,1}[I\|i]{1,1}[T\|t]{1,1}\s$/); #Easier to match /quit/i next if ($_ =~ /^$/ or $_ =~ /^\s+$/); if($line =~ /\s([a-zA-Z]+\s)+\s(-?\d\d?)\s$/) { print $_; } }` [download] Now - if you supply a string of "this that other 555" it will NOT match, because it matches 2 digits, but fails on the trailing "$" in the regex. "By three methods we may learn wisdom: First, by reflection, which is noblest; Second, by imitation, which is easiest; and third by experience, which is the bitterest." -Confucius	[reply] [d/l]
Re: New to perl and regex by Athanasius (Archbishop) on Nov 19, 2012 at 02:15 UTC
`if($line =~ /\s([a-zA-Z]+\s)+\s(-?\d\d?)\s/) ^` [download] The problem description specifies “at least 1 or more white space” between the last word and the integer, but the regex has `\s` (which is zero or more). This should be `\s+` (one or more). Apart from that, it’s difficult to offer advice without more information. Specifically, how is the code failing? As Kenosis says, please provide some sample data (along with the desired output). Athanasius <°(((>< contra mundum*	[reply] [d/l]
Re: New to perl and regex by Kenosis (Priest) on Nov 19, 2012 at 01:37 UTC
Hi, MarvC, and welcome to PerlMonks! Do you have a sample string you can share that you're trying to match? Also, help me understand what ...words saved in a buffer... means.	[reply]
Re: New to perl and regex by Utilitarian (Vicar) on Nov 19, 2012 at 08:44 UTC
Neatened up your regular expressions a bit: `print "Enter a string (enter quit or [CTRL]-D to quit): "; while(<>) { $line = $_; last if ($_ =~ /^\squit\s$/i); next if ($_ =~ /^\s~$/); if($line =~ /\s[a-zA-Z\s]+\s-?\d+\s/) { print $_; } }` [download] `print "Good ",qw(night morning afternoon evening)[(localtime)[2]/6]," fellow monks."`	[reply] [d/l] [select]
Re^2: New to perl and regex by rovf (Priest) on Nov 19, 2012 at 09:18 UTC
I think it can be made even simpler than that: `/[A-Za-z\s]+\s-?\d\d?/` [download] -- Ronald Fischer <ynnor@mm.st>	[reply] [d/l]


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