Re: File creation issue

[Conrad Schilbe]

Adam Butler wrote:
I forgot to mention that $file is set via a form.  On first run of the
script, it prints a form to the browser asking for a file name, which you
enter, and then it's submited and sent to the script.  The way I understood
it, when open tries to open a file that doesn't exist, it creates it.
Thanks,
 

The source of "$file" doesn't really matter, but I hope you are 
sanitizing that input!

Open(GAMELOG, "$file");
   @entries = <GAMELOG>;
close(GAMELOG);
     

This snippet will not create the file if it doesn't exist. As mentioned 
before, you can open a file for read or write. When opening for write, 
the file will be created if it does not exist but what would be the 
point of this if you are only reading the file?

If this is actually what you want, add a test routine that creates the 
file if it doesn't exist:

if (!( -f $file )) {
   # create the file
   open(FH, "> $file");
   close(FH);
}
open(GAMELOG, "$file") || die "$!\n";
   @entries = <GAMELOG>;
close(GAMELOG);
If you want to just test for the file instead of creating it too:
if ( -f $file ) {
   open(GAMELOG, "$file") || die "$!\n";   # you may still die with 
permission problems even though it exists.
      @entries = <GAMELOG>;
   close(GAMELOG);
}

-- cs
You've had your answer. in your script 'Open' means nothing and the
file won't be created even if you use 'open'.
      #!/usr/bin/perl
      chdir "/tmp";
      $log = "game.log";
      open LOG, ">$log" or die $!; # --->   >  !
      print LOG "success !";
      close LOG;
      open LOG, $log;
      for (<LOG>) { print };
Make it a rule NEVER to open a filehandle without testing.
JD