Urgent

From: Lktee (lktee_at_bantex.com.my)
Date: 05/29/04 

To: "'John W. Krahn'" <krahnj@acm.org>
Date: Sat, 29 May 2004 13:15:12 +0800

Other command can instance? Because after try still no work.

-----Original Message-----
From: John W. Krahn [mailto:krahnj@acm.org]
Sent: Thursday, May 27, 2004 9:37 AM
To: beginners@perl.org
Subject: Re: How to find same filename in subdirectory

 
> Hi, John thanks you comment, the coding is work. But i'm detect have some
small
> problem in line 7 print "$File::Find::name\n" if $_ eq 'File1'; I test as
below
> sample 2 is follow you coding. From here i detect nothing return from
script.
>
> But after i amend the coding become print "$File::Find::name\n" if $_ =
'File1';
> this work and return the result i expected. But why perl show "Found = in
> conditional, should be == at ./callback line 7."? Any idea? I try put "=="
to
> replace "=", the result return like sample 2.

Perhaps you want this instead:

print "$File::Find::name\n" if /File1$/;

John 

-- 
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