s/\.0+(\d+)/\.\1/;
The above substitution works.
Ashok
On 11/13/07, danlamb <[EMAIL PROTECTED]> wrote:
>
> Hey guys,
>
> I need a regex to remove zeros after the . in a file name if they are
> followed by another digit.
>
> Example: XX.001 becomes XX.1
>
> Any ideas? I've been banging
Or
You can also do ...
$string =~ s/(\d+)$/$1+1/e;
Ashok
On 10/10/07, Rob Dixon <[EMAIL PROTECTED]> wrote:
>
> Tatiana Lloret Iglesias wrote:
> >
> > Hi all!
> >
> > What regular expression do I need to convert Version: 1.2.3 to
> Version:
> > 1.2.4 ?
> >
> > I.e. my pattern is Version: num
Any clue from anyone ?.
Am still struggling with the problem.
Ashok
On 2/1/07, Ashok Varma <[EMAIL PROTECTED]> wrote:
Hi,
I have XSD, XML files and want to validate the XML against the XSD and
then only proceed with parsing the XML file.
I used XML::SAX::ParserFactory, XML::Val
Hi,
I have XSD, XML files and want to validate the XML against the XSD and then
only proceed with parsing the XML file.
I used XML::SAX::ParserFactory, XML::Validator::Schema modules but following
is the issues i faced ...
1. Failed with this reason --- elementFormDefault in must be
'unqualifie
Hi Michael,
Following is one way ...
Here is am assuming that you have dumped that data into a array named
'@your_array' which contains each line in array references(array of arrays).
$table = new HTML::T
Hi Kishore,
The below snippet will get your desired result.
---
#!/usr/bin/perl
use strict ;
my @structure_name = ();
my $fni = 'd:\Sample.txt' ;
my $fno = 'd:\ashok.txt'
If you can provide me a sample file maybe i can help more.
i am guessing your file and writing a sample below.
-
open(FH, "/your/file/path");
my @file = ;
close FH;
open(FH, ">/your/new/file");
foreach my $line (@file)
