如何在PHP中获取当前包含脚本的URL？

Question

I have a file called index.php which includes a class from another file (class/event.class.php) with the following code:

 <?php
      require_once('class/event.class.php');
      $evt = new Event;
      $evt->doSomething();
 ?>

The file event.class.php should be able to read the URL used in the request (not $_SERVER["SCRIPT_FILENAME") to the file.

ie: http://localhost:8888/class/event.class.php

How do I do that?

Answer 1

Using the $_SERVER array, example:

$protocol = 'http';
$port = '';

if ( isset( $_SERVER['HTTPS'] ) && strcasecmp( $_SERVER['HTTPS'], 'on' ) == 0 ) {
    $protocol = 'https';
}
if ( isset( $_SERVER['SERVER_PORT'] ) ) {
    $port = ':' . $_SERVER['SERVER_PORT'];
}

// Don't display the standard ports :80 for http and :443 for https
if ( ( $protocol == 'http' && $port == ':80' ) || ( $protocol == 'https' && $port == ':443' ) ) {
    $port = '';
}

$host_url = $protocol . '://' . $_SERVER['SERVER_NAME'] . $port;
$base_url = rtrim( str_replace( basename( $_SERVER['SCRIPT_NAME'] ), '', $_SERVER['SCRIPT_NAME'] ), '/' );