I have the following code:
if (include_once(dirname(__FILE__).'/file.php')
|| include_once(dirname(__FILE__).'/local/file.php')
)
{
This causes an error because PHP tries to include "1"
(presumably dirname(__FILE__).'/file.php' || dirname(__FILE__).'/local/file.php'
)
Commenting out the second line makes this work as intended, other than the fact it won't use the second file. Do I really need to use elseif
and code duplication here, or is there a way to get this working?
$ php -开发者_开发百科-version
PHP 5.2.6-3ubuntu4.2 with Suhosin-Patch 0.9.6.2 (cli) (built: Aug 21 2009 19:14:44) Copyright (c) 1997-2008 The PHP Group Zend Engine v2.2.0, Copyright (c) 1998-2008 Zend Technologies
Group the include
statements:
if ( (include_once dirname(__FILE__).'/file.php')
||
(include_once dirname(__FILE__).'/local/file.php')
)
See example #4 in the manual page:
<?php
// won't work, evaluated as include(('vars.php') == 'OK'), i.e. include('')
if (include('vars.php') == 'OK') {
echo 'OK';
}
// works
if ((include 'vars.php') == 'OK') {
echo 'OK';
}
?>
Try using parentheses to subdue the operator precedence.
if ( (include_once('xml.php')) || (include_once('xml.php')) ) {
echo 'bah';
}
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