Get Initial php script file name not by FILE

Question

I'm looking for a way to get what the current script running. But the thing is I can't use __FILE__ since it gives me the file i'm currently in and in the case of an included file, I want the initial file.

So if I run index.php and in that file there is an include('test.php') I want to be able in test.php to get 'index.php' as the current file im runing.

I know I could do __FILE__ in index.php and keep the value in an variable. But i'm looking for a way with minimal change to my files.

Also the script is running in CLI on windows. So trying to parse the the path wont work also.

What are my options?

Thanks to everyone taking the time to read this :)

Everything's inside $_SERVER.

Álvaro González
– Álvaro González

2016-02-10 16:14:04 +00:00
Commented Feb 10, 2016 at 16:14 — Álvaro González
– Álvaro González, Commented Feb 10, 2016 at 16:14
... or $argv[0] because CLI ...

Joe Watkins
– Joe Watkins

2016-02-10 16:17:42 +00:00
Commented Feb 10, 2016 at 16:17 — Joe Watkins
– Joe Watkins, Commented Feb 10, 2016 at 16:17

fusion3k · Accepted Answer · 2016-02-10 16:24:52Z

6

You have to use $_SERVER[PHP_SELF].

According to documentation:

The filename of the currently executing script, relative to the document root.

$_SERVER[PHP_SELF]works also on CLI interface. (Edit: ... but I doesn't have tested in Windows)

answered Feb 10, 2016 at 16:24

fusion3k

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2 Comments

Jaxx0rr Over a year ago

Fatal error: Uncaught Error: Undefined constant "PHP_SELF"

izeke Over a year ago

PHP_SELF needs to be in quotes - otherwise it will be parsed as a constant

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