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

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"

其他回答

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

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"
?>

您可以使用basename()函数。