php realpath 不产生可用的结果

发布于 2024-09-09 02:17:50 字数 471 浏览 1 评论 0原文

<?php
    $server = $_SERVER["SERVER_NAME"];
    $pathpath = realpath("../../files/uploaded_file.jpg");
    echo "You can link to the file using the following link... $server$pathpath";
?>

不幸的是，这会产生以下结果...

www.example.com/home/fhlinux123/g/example.com/user/htdocs/ninja/base/files/1.doc

而我所追求的结果如下...

www.example.com/files/uploaded_file.jpg

我不能假设文件夹“文件”将始终位于同一目录中。

原文

<?php
    $server = $_SERVER["SERVER_NAME"];
    $pathpath = realpath("../../files/uploaded_file.jpg");
    echo "You can link to the file using the following link... $server$pathpath";
?>

Unfortunately this produces the following...

www.example.com/home/fhlinux123/g/example.com/user/htdocs/ninja/base/files/1.doc

Whereas what I am after is as follows...

www.example.com/files/uploaded_file.jpg

I can't assume that the folder 'files' will always be in the same directory.

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蓝天白云 2024-09-16 02:17:50

这是因为 realpath 返回服务器框根的绝对路径，而不是 Web 服务器 htdocs 根的绝对路径。
您可以从 $_SERVER["DOCUMENT_ROOT"] 检索网络服务器 htdocs 根目录，然后从 realpath 返回的结果的开头将其删除

简单而肮脏的示例：

$server = $_SERVER["SERVER_NAME"]; 
$pathpath = realpath("../../files/uploaded_file.jpg"); 
$serverPath = $_SERVER["DOCUMENT_ROOT"];
$pathpath = substr($pathpath,strlen($serverPath));

echo "You can link to the file using the following link... $server$pathpath";

That's because realpath returns the absolute path from the server box root, not the webserver htdocs root.
You can retrieve the webserver htdocs root from $_SERVER["DOCUMENT_ROOT"], then strip that from the beginning of the result returned by realpath

Quick and dirty example:

$server = $_SERVER["SERVER_NAME"]; 
$pathpath = realpath("../../files/uploaded_file.jpg"); 
$serverPath = $_SERVER["DOCUMENT_ROOT"];
$pathpath = substr($pathpath,strlen($serverPath));

echo "You can link to the file using the following link... $server$pathpath";

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