How to compare two file structures in PHP?
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by OM The Eternity
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Published on 2010-05-06T04:51:38Z
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2010/05/06
4:58 UTC
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I have a function which gives me the complete file structure upto n-level,
function getDirectory($path = '.', $ignore = '') {
$dirTree = array ();
$dirTreeTemp = array ();
$ignore[] = '.';
$ignore[] = '..';
$dh = @opendir($path);
while (false !== ($file = readdir($dh))) {
if (!in_array($file, $ignore)) {
if (!is_dir("$path/$file")) {
//display of file and directory name with their modification time
$stat = stat("$path/$file");
$statdir = stat("$path");
$dirTree["$path"][] = $file. " === ".
date('Y-m-d H:i:s', $stat['mtime']) . " Directory ==
".$path."===". date('Y-m-d H:i:s', $statdir['mtime']) ;
} else {
$dirTreeTemp = getDirectory("$path/$file", $ignore);
if (is_array($dirTreeTemp))$dirTree =
array_merge($dirTree, $dirTreeTemp);
}
}
}
closedir($dh);
return $dirTree;
}
$ignore = array('.htaccess', 'error_log', 'cgi-bin', 'php.ini', '.ftpquota');
//function call
$dirTree = getDirectory('.', $ignore);
//file structure array print
print_r($dirTree);
Now here my requirement is , I have two sites
- The Development/Test Site- where i do testing of all the changes
- The Production Site- where I finally post all the changes as per test in development site
Now, for example, I have tested an image upload in the Development/test site, and i found it appropriate to publish on Production site then i will completely transfer the Development/Test DB detail to Production DB, but now I want to compare the files structure as well to transfer the corresponding image file to Production folder.
There could be the situation when I update the image by editing the image and upload it with same name, now in this case the image file would be already present there, which will restrict the use of "file_exist" logic, so for these type of situations....HOW CAN I COMPARE THE TWO FILE STRUCTURE TO GET THE SYNCHRONIZATION DONE AS PER REQUIREMENT??
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