如果我有PHP脚本,我怎么能得到当前执行的文件没有其扩展名的文件名?

给定一个“jquery.js.php”形式的脚本的名称,我如何提取只是“jquery.js”部分?


当前回答

例子:

包含文件:config.php

<?php
  $file_name_one = basename($_SERVER['SCRIPT_FILENAME'], '.php');
  $file_name_two = basename(__FILE__, '.php');
?>

执行文件:index.php

<?php
  require('config.php');
  print $file_name_one."<br>\n"; // Result: index
  print $file_name_two."<br>\n"; // Result: config
?>

其他回答

你也可以用这个:

echo $pageName = basename($_SERVER['SCRIPT_NAME']);

试试这个

$current_file_name = $_SERVER['PHP_SELF'];
echo $current_file_name;

以下是我最近在寻找答案时发现的一个列表:

//self name with file extension
echo basename(__FILE__) . '<br>';
//self name without file extension
echo basename(__FILE__, '.php') . '<br>';
//self full url with file extension
echo __FILE__ . '<br>';

//parent file parent folder name
echo basename($_SERVER["REQUEST_URI"]) . '<br>';
//parent file parent folder name with //s
echo $_SERVER["REQUEST_URI"] . '<br>';

// parent file name without file extension
echo basename($_SERVER['PHP_SELF'], ".php") . '<br>';
// parent file name with file extension
echo basename($_SERVER['PHP_SELF']) . '<br>';
// parent file relative url with file etension
echo $_SERVER['PHP_SELF'] . '<br>';

// parent file name without file extension
echo basename($_SERVER["SCRIPT_FILENAME"], '.php') . '<br>';
// parent file name with file extension
echo basename($_SERVER["SCRIPT_FILENAME"]) . '<br>';
// parent file full url with file extension
echo $_SERVER["SCRIPT_FILENAME"] . '<br>';

//self name without file extension
echo pathinfo(__FILE__, PATHINFO_FILENAME) . '<br>';
//self file extension
echo pathinfo(__FILE__, PATHINFO_EXTENSION) . '<br>';

// parent file name with file extension
echo basename($_SERVER['SCRIPT_NAME']);

别忘了删除:)

< br >

试试这个

$file = basename($_SERVER['PATH_INFO'])

虽然__FILE__和$_SERVER是最好的方法,但在某些情况下,这也是一种替代方法:

get_included_files();

它包含调用它的文件路径和所有其他包含。