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Editorial Team
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Asked: 2026-06-16T09:27:35+00:00

I am trying to include a php file from the parent directory and I

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I am trying to include a php file from the parent directory and I getting error:

admin@webby:~$ /usr/bin/php
/var/phpscripts/email_parser/tests/_email-test.php PHP Fatal error:
require_once(): Failed opening required ‘../PlancakeEmailParser.php’
(include_path=’.:/usr/share/php:/usr/share/pear’) in
/var/phpscripts/email_parser/tests/_email-test.php on line 6

Fatal error: require_once(): Failed opening required
‘../PlancakeEmailParser.php’
(include_path=’.:/usr/share/php:/usr/share/pear’) in
/var/phpscripts/email_parser/tests/_email-test.php on line 6

PHP file

#!/usr/bin/php
<?php
error_reporting(E_ALL ^ E_NOTICE ^E_WARNING);
ini_set("display_errors" , 1);

require_once("../PlancakeEmailParser.php");

// etc

?>

Folder Structure

admin@webby:/var/phpscripts/email_parser$ find .
.
./composer.json
./README.txt
./LICENSE.txt
./PlancakeEmailParser.php
./tests
./tests/_email-test.php

For testing it works fine when I move PlancakeEmailParser.php into the tests directory and remove the “../” from the require

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1 Answer

  1. Editorial Team

    The line

    require_once("../PlancakeEmailParser.php");

    Is not using a fully qualified file-system path or PHP-URI-scheme. Therefore its outcome depends on PHP configuration, most often because of the include directory configuration.

    The PHP CLI can use a different configuration file than with your webserver – or – the working directory is different as with your webserver (probably the later plays more of a role in your specific scenario).

    What you want can be easily expressed fully qualified, too:

    require_once(__DIR__ . "/../PlancakeEmailParser.php");

    This is probably what you’re looking for.

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