Single source for my debug script so __FILE__ still works

[appobs]

I have this at top of my index page:

 

<?php

$debug = 'off';

define (SCR_PATH, '/home/buildmyb/scripty');

// Obligatory config file

include SCR_PATH . '/config.php';

// tools

include SCR_PATH . '/tools.php';

?>

and this debug thing in each file that gets included:

 

<?php

$debug = ''; // initialise it so if anything other then empty, do NOT do the if

if ($debug === '') {

    echo 'this line is in ' . __FILE__;

    echo '<hr>';

}

?>

I'd like a single source for the debug thing but, obviously I can't include() it in each include or __FILE__ will show info for that, not the one I want.

 

 

What do I do?

[Jacques1]

You can use debug_backtrace() to get the original file which included your debug script. It's the last entry in the list.

Edited September 25, 2015 by Jacques1

[appobs]

Thanks, looking now

[appobs]

You can use debug_backtrace() to get the original file which included your debug script. It's the last entry in the list.

 

Nice one

 

Test files set up as below, will throw something similar in my live app in a sec.

 

 index.php

<h4>This line HTML in index.php</h4>
<?php
include 'config.php';
include 'tools.php';
?>

config.php and tools.php

<h4>This line HTML in config.php</h4>
<?php
include 'debug.php';
?>

<h4>This line HTML in tools.php</h4>
<?php
include 'debug.php';
?>

debug.php

<h4>This line HTML in debug.php</h4>
<?php
$array = debug_backtrace(DEBUG_BACKTRACE_IGNORE_ARGS);
echo basename($array['0']['file']) . ' is loaded<br>';
?>