PHP include with ../ failed open - php

Question

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

Answer 1

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.

Answer 2

Quote from PHP CLI SAPI: Differences to other SAPIs:
It does not change the working directory to that of the script. (-C
and --no-chdir switches kept for compatibility)
In order to keep relative paths in your script working as-is, change directory to where the script resides, then execute the script:
cd /var/phpscripts/email_parser/tests/ && ./_email-test.php