我有一个url字符串,如下所示:
http://example.com/sfm?dir=uploads/sfm/root/folder5/file.zip
echo $_GET['extract']
输出:uploads/sfm/root/folder5/file.zip
我如何剥离最后一个文件.zip,以便
echo $_GET['extract']
输出:uploads/sfm/root/folder5/
使用以下代码:
$String = $_GET['extract'];
$Words = explode('/', $String);
echo end($Words );
只需使用目录名-返回父目录的路径
echo dirname($_GET['extract'])."/";
尝试:
$url = "uploads/sfm/root/folder5/file.zip";
$urlArr= explode("/",$url);
array_pop($urlArr);
echo implode("/",$urlArr);
也许这有效:
echo substr((string)$_GET['extract'],26);
尝试这个
<?php
$ss ="uploads/sfm/root/folder5/file.zip";
$vv = explode('/',$ss);
array_pop($vv);
$mm =implode('/',$vv);
print_r($mm);
?>