Php - get parent-script name

问题

parent.php:

require_once 'child.php';

child.php:

echo __FILE__;

It will show '.../child.php'

How can i get '.../parent.php'

回答1:

print $_SERVER["SCRIPT_FILENAME"];

回答2:

The chosen answer only works in environments that set server variables and specifically won’t work from a CLI script. Furthermore, it doesn't determine the parent, but only the topmost script file.

You can do almost the same thing from a CLI script by looking at $argv[0], but that doesn’t provide the full path.

The environment-independent solution uses debug_backtrace:

function get_topmost_script() {
  $backtrace = debug_backtrace(
      defined("DEBUG_BACKTRACE_IGNORE_ARGS")
      ? DEBUG_BACKTRACE_IGNORE_ARGS
      : FALSE);
  $top_frame = array_pop($backtrace);
  return $top_frame['file'];
}

回答3:

I don't think you can do that : the __FILE__ magic constant indicates in which file it is written ; and that is all.

If you want to know which PHP script was initially called (which URL was requested, for instance), you might have more luck looking at the $_SERVER superglobal : it contains many informations, including some that will help you (like SCRIPT_FILENAME or SCRIPT_NAME, for instance) ;-)

回答4:

Straight to the solution:

Parent Script:

echo $_SERVER['SCRIPT_FILENAME'];

Child Script (included file in parent script):

echo __FILE__;

Make a file called "parent.php" with contents:

include('child.php');

Make a file called "child.php" with contents:

echo "My Parent is at: " . $_SERVER['SCRIPT_FILENAME'] . "<br>";
echo "I'm at: " .  __FILE__;

You get the idea...

回答5:

You have to use basename($_SERVER["SCRIPT_FILENAME"]) or, if you like the script nam only, you can use basename($_SERVER["SCRIPT_FILENAME"], '.php').

来源：https://stackoverflow.com/questions/1318608/php-get-parent-script-name

标签

php

parent