Download files from url of site directory - php

Question

Two zip files and Two images are in https://example.com/directory. (Since index.php file is available in this directory we don't know the file names of those files).
Is there any possible way to find names of above four files using PHP or Linux commands?

Answer 1

You must keep in mind: if your web-server disallows to scan a directory then you couldn't get file names. But if the directory is shared in web-server you can get a list of files with using wget command. For example:
wget -nv -r -np '*.*' https://example.com/directory/

Answer 2

Say you have on your webserver example.com a directory like: /multimedia/ and you want to retrieve its contents like mp3, jpg, png, etc - but not the listing index nor tmp index files like i.e: index.html•C=D;O=D:
wget -r -R "*.html*" -np -nd -l 1 example.com/multimedia/
where:
-r # Recursive
-R "*.html*" # Reject tmp and listing html files
-np # Don't ascend to the parent directory
-nd # Don't create parent directories
-l 1 # Only one level deep (multimedia/ folder only)
To learn more: run man wget or wget --help or visit gnu.org WGET

Answer 3

<?php
$files=[];
$folder="/directory/directory/directory/";
if (is_dir($folder)){ // check whether exists or not
$allfiles = glob($folder.'/*.*'); // Read all filenames of $folder directory
foreach($allfiles as $imgFile) {
$files[]=$imgFile; // Putting the names of $folder directory to array one by one
}
}
echo "<pre>";
print_r($files); // printing the file name array to list aal names
echo "</pre>";

Answer 4

I think you don't understant HTTP protocals at all. According to HTTP protocal it doesn't know about any directory/sub-directory. You can't do things as such in HTTP Protocal.
You want to download http://example.com/directory some files for which you don't know the names. For that you can't use PHP as such. You have to depend on external library such as wget etc. or FTP protocal.

Answer 5

<?php
$dir = 'directory';
$filesArr = scandir($dir);
$files = array();
for($i=2; $i<count($filesArr);++$i){
if($filesArr[$i]=="index.php"){ continue; }
$files[] = $filesArr[$i];
}
$zipname = 'file1.zip';
$zip = new ZipArchive;
$zip->open($zipname, ZipArchive::CREATE);
foreach ($files as $file) {
$zip->addFile($file);
$zip->addFile($dir.'/'.$file, $dir.'/'.$file);
}
header('Content-Type: application/zip');
header('Content-disposition: attachment; filename='.$zipname);
header('Content-Length: ' . filesize($zipname));
readfile($zipname);
$zip->close();
exit;
?>

Answer 6

Try this I am using below sample in my project and its working fine for me,
<?php
$directory='directory';
$handler= opendir("$directory"."/");
$i=0;
while (false !== ($file = readdir($handler))) {
if ($file != '.' && $file !='..'){
echo $file.'<br>';
}
}
?>

Answer 7

You can use PHPs scandir Method.
Try this:
<?php
$files = scandir('./');
print_r($files);
This should print all files in the current directory. More information here