Regex with digits

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

Hello Monks , I have this part :

if ($lastOne =~ /\w+$/ && $lastOne !~ /(?<=\W)[0-9]$/)
[download]

to check for lines end with \digit , I want to see if the line end with a back slash follow by a digit then I want to ignore it . the data i have is the following :

hull/uuu/8
cook/too/un-8-9
luck/goo
[download]

so in this data I want to ignore only the first line , my code is ignoring both the first and second . do you see the problem .. thanks

Comment on Regex with digits Select or Download Code

Replies are listed 'Best First'.
Re: Regex with digits by Abigail-II (Bishop) on Jul 31, 2002 at 14:03 UTC
`next if m{/\d$};` [download] Abigail	[reply] [d/l]
Re: Regex with digits by mkmcconn (Chaplain) on Jul 31, 2002 at 17:47 UTC
Abigail-II's solution filters out all lines that end in a single-digit, which is exactly++ what you asked for here. But, if you are also trying to filter out strings with 'non-word' characters in the final chunk as you sought in this thread, remember that '\w' matches only alphanumeric plus '_'. It will not match '-' so, if you are still trying to perform both filters, you will continue to skip line 2 of your example. Combining the two requirements, adding '-' to the 'words' character set, does this do what you want? `#!/usr/bin/perl -w use strict; foreach my $lastone (<DATA>){ if ($lastone =~ m{/[-\w]+$} && $lastone !~ m{/\d$}){ print $lastone; } } __END__ hull/uuu/8 cook/too/un-8-9 luck/goo@@ luck/goo yty~uyuoo~iu99~##uyt~hu/dwo9 chuck/ii@@/a_` [download] mkmcconn	[reply] [d/l]
Re: Regex with digits by robobunny (Friar) on Jul 31, 2002 at 14:50 UTC
at the risk of being nitpicky, the character that you are looking for, specifically '/', is a slash, not a backslash.	[reply]


Pathologically Eclectic Rubbish Lister
	PerlMonks