why do i get 1 instead of the file required? [duplicate]

This quest开发者_开发知识库ion already has answers here: Closed 11 years ago.

Possible Duplicate:

including php file from another server with php

i want to have my own error system instead of having the php errors so for example i require a file from another server and that server is not available right now

<?php
require 'http://example.com/file.php' or die ("that host is not available right now");
?>

but instead i get

Warning: require(1) [function.require]: failed to open stream: No such file or directory in C:\xampp\htdocs\index.php on line 5

Fatal error: require() [function.require]: Failed opening required '1' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\index.php on line 5

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It's because require 'foo' or bar() is interpreted as require ('foo' or bar()). 'foo' or bar() equals true, i.e. 1. If you want to write it like this, use different parentheses:

(require 'http://example.com/file.php') or die ("that host is not available right now");

But, you don't need die at all here, since require will already halt program execution if the required file can't be loaded. Just require 'http://example.com/file.php'; will do fine. Whether you should actually load foreign PHP files over the network is another story (hint: probably not).

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The problem is the precedence of operators. The PHP are including the result of the "or" comparison (true). Try to remove this.

include('http://...') or die('error...');

It will work.