I think the implication of the oringal poster was not the include/requires
stuff,
but more a build of PHP with --enable-1000000 options vs php-shared....etc



-----Original Message-----
From: Maxim Maletsky (PHPBeginner.com)
[mailto:[EMAIL PROTECTED]]
Sent: Friday, 5 October 2001 11:33 AM
To: 'Opec Kemp'; [EMAIL PROTECTED]
Subject: RE: [PHP] PHP is a Bloated Kludge, discuss!


> > PHP on the other hand seems to load in ALL the code that
> MAY be run.
> > i.e. an include brings things in which are inside an if,
> even if the
> > if equates to false.
>
> A quick check in the manual, it states that:
>
> "Unlike include(), require() will always read in the target
> file, even if the line it's on never executes. If you want to
> conditionally include a file, use include(). The conditional
> statement won't affect the require(). However, if the line on
> which the require() occurs is not executed, neither will any
> of the code in the target file be executed. "
> http://www.php.net/manual/en/function.require.php
> and
> http://www.php.net/manual/en/function.require-once.php
>
>
> So to do what you said i.e. conditionally load the
> function/files that you want then you should use include().
> http://www.php.net/manual/en/function.include.php
> and
> http://www.php.net/manual/en/function.include-once.php
>
> So I don't buy this agument.

This is outdated. Look int othe archives - I had this discussion with
Zeev close to a year ago - since PHP 4.0.(something) require and include
behaves in the exactly same way.

Test it yourself on the latest release of PHP:


  require.test.php
 ----
 <?

 echo 'included?';

 if($go)
        require 'require.error.test';

 ?>
 ----

  require.error.test
 ----
 <?

 echo '<H1>yup!</H1>';

 parse error;

 ?>
 ----

Access it with:
 1. require.test.php?go=e

And then try it with:
 2. require.test.php


What do you buy now?


Maxim Maletsky
www.PHPBeginner.com


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