hey, do you have a solution to this?
I need to upload a file, but i just want the file name, not the entire
path. Currently, it saves the actual file on the server as:
G:\ISIR\ISIR2002gallery\images\isir1.jpg as the FILENAME!!!
i just want it to save as "isir1.jpg" or whatever the filename is
On Sat, 21 Dec 2002 14:07:03 -0700, [EMAIL PROTECTED] (Eric Lin)
wrote:
> Now I am in linux again,
>I did /usr/bin/perl /usr/lib/cgi-bin/showphoto.pl > mytestout.jpg
>then at netscape 7 's url type /home/fsshl/mytestout.jpg
>
>it showed
>---
>The image "file:///home/fssh
On Sat, 21 Dec 2002 20:32:43 EST, [EMAIL PROTECTED] wrote:
>Is there a way to have my script create a list of all the directories and
>files on my server from my root directory down?
It will create a huge file. Since there is a c program to do this,
you are better off using "find".
"find / -
Use regular expressions to get only the file name from the path.
You should use something like this:
$path =~ s/^.*[\\\/]//;
Teddy,
Teddy's Center: http://teddy.fcc.ro/
Email: [EMAIL PROTECTED]
- Original Message -
From: "james lundeen" <[EMAIL PROTECTED]>
To: "beginners-cgi" <[EMAIL P
Holiday greetings,
I would like to be able to test for either of the two secret words
but it seems to fail, what am I missing?
Thanks
Dave
#!/usr/bin/perl
use CGI qw/:standard/;
use CGI::Carp qw(fatalsToBrowser);
use strict;
my $qs = $ENV{'QUERY_STRING'};
my $secret_word = 'one';
my $secret
