hi

this is what i tried

#!/sbcimp/run/pd/perl/5.8.2/bin/perl -w
my $name = basename($FILENAME, 
"\Developer\view_local\local_nt\FDAFDSAFDSASDFA\ASDFDAFSASDF\NewProcess_date_22-oct-2004.log");
$name =~ s/_*.*//s;
print $name;

does it make sence

Regards
Sreedhar

-----Original Message-----
From: Chris Devers [mailto:[EMAIL PROTECTED]
Sent: 22 October 2004 16:52
To: Kalkunte-Venkatachala, Sreedhar
Cc: Perl Beginners List
Subject: RE: how the print the first 7 letter of file name


On Fri, 22 Oct 2004 [EMAIL PROTECTED] wrote:

> I have written a program to monitor logs. Here I am substituting the 
> file name dynamically say : 
> /irrelevant_path_info/fadsdsafasdffdsar_asdffdsa_asfdfsad_asfdfsda.log
> 
> would be: _fadsds
> 
> So, I am not getting logic to capture this name
> 
> Hope I am clear of my explanation

You're getting there, but you've still shown no code for this. 

You seem to want the first 7 characters after the last slash. 

This can be done with one line, but I'm not going to give it to you.

You have to demonstrate that you've at least attempted to write it.

Only then will I show you how I might try to solve this problem.



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Chris Devers

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