Small code, newbie to PHP

[fqservers]

Dear users,

 

I have the following function

 

<? 
$page = 0; 
$URL = "http://www.xxxxxxx.com/yyyyy.php/"; 
$page = @fopen($URL, "r");

 

And the function continues. I would like to exit the function if the above http://www.xxxxxxx.com/yyyyy.php/ does not exit. Because right now when i execite it for a non existing .php file it crashes my computer. I would really appreciate any help. Thank you.

 

P.S I have very little knowledge of PHP

 

 

 

 

[Ken2k7]

Try

<?php
$page = 0; 
$URL = "http://www.xxxxxxx.com/yyyyy.php/"; 
if (!fopen($URL, "r")) exit();
$page = @fopen($URL, "r");
?>

[fqservers]

Try

<?php
$page = 0; 
$URL = "http://www.xxxxxxx.com/yyyyy.php/"; 
if (!fopen($URL, "r")) exit();
$page = @fopen($URL, "r");
?>

 

OMG it worked. Thank you so much:

 

now rather than crashing my computer it displays the follwoing message.

 

Warning: fopen(http://www.xxxxxxx.com/yyyyy.php/) [function.fopen]: failed to open stream: HTTP request failed! HTTP/1.1 404 Not Found in /home/jessy/public_html/yyyyy.php on line 74

 

is it possible to alter this error message?

 

thank you for your help again.

[Ken2k7]

Define alter this message.

 

Does xxxxxxx.com and yyyyy.php exist?

[fqservers]

Define alter this message.

 

Does xxxxxxx.com and yyyyy.php exist?

 

yes they do exit and some times it is possible that they are not online.

 

How can I define alter this message? I have very little experience in PHP I would reall apreciate if you can provide me more detail.

 

thank you

[trq]

<?php

  $URL = "http://www.xxxxxxx.com/yyyyy.php/"; 
  if (!$page = @fopen($URL, "r")) {
    exit();
  }

?>

[abdfahim]

there is know need for using extra

$page = @fopen($URL, "r");

, because when you write

if (!fopen($URL, "r")) exit();

, it means the program will exit if it can't open file or it'll open the file if it can.

[fqservers]

Thank you all for your help. You solve my major problem

but I still get an error message

 

"

Warning: fopen(http://www.xxxxxxx.com/yyyyy.php/) [function.fopen]: failed to open stream: HTTP request failed! HTTP/1.1 404 Not Found in /home/jessy/public_html/yyyyy.php on line 74" I would like to have a custom error message. But this is minor

 

thankx again

[asmith]

did you use the thorpe code ?

that should work fine ,

 

however if you still have it, post your final code .