Create Zip file issue

sss.ark · Joined: 28 Jan 2010 Posts: 2

hi,
i have to convert folder to zip.

my script path is www.domainname.com/create_zip.php
Folder location : www.domainname.com/Main_dir/12343/sample/test.jpg.
i wants to zip the sample folder alone.

if i create means it will come along with the root directory (Main_dir/12343/sample)

i need only sample folder content to be zipped. no trace for parent directory.

how to proceed.
thanks in advance.

qforever · Joined: 01 Feb 2010 Posts: 4

I made this using zip.lib.php(it goes with http://www.phpmyadmin.net) but I zipped from strings(just fread their contents and gaev them names directly).
But I'm not sure that this advice will help you. Please show your code.

sss.ark · Joined: 28 Jan 2010 Posts: 2

hi,
thanks for ur response.

i found another solution.
find below.

$dirName = $path;

if (!is_dir($dirName)) {
throw new Exception('Directory ' . $dirName . ' does not exist');
}

$dirName = realpath($dirName);
if (substr($dirName, -1) != '/') {
$dirName.= '/';
}
$dirStack = array($dirName);
$cutFrom = strrpos(substr($dirName, 0, -1), '/')+1;
while (!empty($dirStack)) {
echo $currentDir = array_pop($dirStack);
echo "<br>";
$filesToAdd = array();

$dir = dir($currentDir);
while (false !== ($node = $dir->read())) {
if (($node == '..') || ($node == '.')) {
continue;
}
if (is_dir($currentDir . $node)) {
array_push($dirStack, $currentDir . $node . '/');
}
if (is_file($currentDir . $node)) {
$filesToAdd[] = $node;
}
}

echo $localDir = substr($currentDir, $cutFrom);
echo "<br>";
$zip->addEmptyDir($localDir);

foreach ($filesToAdd as $file) {
$zip->addFile($currentDir . $file, $localDir . $file);
}
}