basename函数应该给你你想要的:

给定一个包含路径的字符串 文件，此函数将返回 文件的基本名称。

例如，引用手册的页面:

<?php
    $path = "/home/httpd/html/index.php";
    $file = basename($path);         // $file is set to "index.php"
    $file = basename($path, ".php"); // $file is set to "index"
?>

或者，在你的情况下:

$full = 'F:\Program Files\SSH Communications Security\SSH Secure Shell\Output.map';
var_dump(basename($full));

你会得到:

string(10) "Output.map"

有几种方法可以获取文件名和扩展名。你可以用下面这个简单易用的。

$url = 'http://www.nepaltraveldoor.com/images/trekking/nepal/annapurna-region/Annapurna-region-trekking.jpg';
$file = file_get_contents($url); // To get file
$name = basename($url); // To get file name
$ext = pathinfo($url, PATHINFO_EXTENSION); // To get extension
$name2 =pathinfo($url, PATHINFO_FILENAME); // File name without extension

您正在寻找basename。

下面的例子来自PHP手册:

<?php
$path = "/home/httpd/html/index.php";
$file = basename($path);         // $file is set to "index.php"
$file = basename($path, ".php"); // $file is set to "index"
?>

试试这个:

echo basename($_SERVER["SCRIPT_FILENAME"], '.php') 

$filename = basename($path);

为了从URI中获得确切的文件名，我将使用以下方法:

<?php
    $file1 =basename("http://localhost/eFEIS/agency_application_form.php?formid=1&task=edit") ;

    //basename($_SERVER['REQUEST_URI']); // Or use this to get the URI dynamically.

    echo $basename = substr($file1, 0, strpos($file1, '?'));
?>