例如,我如何得到output。map

from

F:\程序文件\SSH通信安全\SSH安全Shell\Output.map

使用PHP吗?


当前回答

试试这个:

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

其他回答

<?php

  $windows = "F:\Program Files\SSH Communications Security\SSH Secure Shell\Output.map";

  /* str_replace(find, replace, string, count) */
  $unix    = str_replace("\\", "/", $windows);

  print_r(pathinfo($unix, PATHINFO_BASENAME));

?> 

正文,html, iframe { 宽度:100%; 高度:100%; 溢出:隐藏; } < iframe的src = " https://ideone.com/Rfxd0P " > < / iframe >

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"
$image_path = "F:\Program Files\SSH Communications Security\SSH Secure Shell\Output.map";
$arr = explode('\\',$image_path);
$name = end($arr);

这很简单。例如:

<?php
    function filePath($filePath)
    {
        $fileParts = pathinfo($filePath);

        if (!isset($fileParts['filename']))
        {
            $fileParts['filename'] = substr($fileParts['basename'], 0, strrpos($fileParts['basename'], '.'));
        }
        return $fileParts;
    }

    $filePath = filePath('/www/htdocs/index.html');
    print_r($filePath);
?>

输出将是:

Array
(
    [dirname] => /www/htdocs
    [basename] => index.html
    [extension] => html
    [filename] => index
)

您可以使用basename()函数。