Question

我试图更改文件夹中的每个文件名，例如，如果文件名是style.css而不是我想将其重命名为style_[md5 value of style].css = style_a1b01e734b573fca08eb1a65e6df9a38.css

这是我尝试的内容

if ($handle = opendir("D:/web/htdocs/extra/css/")) {
    while (false !== ($fileName = readdir($handle))) {
        $path_parts = pathinfo($fileName);
        $newName = md5($path_parts['filename']);
        rename($fileName, $newName);
    }
    closedir($handle);
}

我哪里错了？

错误

Access is denied. (code: 5)
The system cannot find the file specified. (code: 2)

Answer 1

不确定在Windows上会发生同样的情况，但是在这里的GNU上......

如果你打印出你打算做什么而不是直截了当地发现你会看到一些缺陷：

rename( ., d41d8cd98f00b204e9800998ecf8427e)                                                                                                                                                     
rename( .., 5058f1af8388633f609cadb75a75dc9d)

例如这样做的：

echo ("rename( ".$fileName.", ".$newName.")\n");

要检查的下一件事可能是更改文件的权利......

Answer 2

// DS to print  \  the split between folder
define('DS',DIRECTORY_SEPARATOR);
// APP_PATH to get application path on the the server
define('APP_PATH',__DIR__.DS);

$oldname = APP_PATH.'css'.DS.'style.css';
/*
when you echo $oldname ,you will get the complete path of file
*/
// check the file is exists or No
if (file_exists($oldname)) {
        $newName = md5($oldname);
        /*add the extension of file that you will rename it */
        rename($oldname, ($newName.'.css'));
}

如何更改文件夹中每个文件的名称

2 个答案: