Question

    $file_name = 'New Folder.zip'
    $zip = new ZipArchive;
    $result = $zip->open($target_path.$file_name);
    if ($result === TRUE) {
        for($i = 0; $i < $zip->numFiles; $i++) {
        $filename = $zip->getNameIndex($i);
        $fileinfo = pathinfo($filename);
        copy("zip://".$file_name."#".$filename, $target_path.$fileinfo['basename']);
        }
    }

When i run this code i get this error Warning: copy(zip://New Folder.zip#New Folder/icon_android.png) [function.copy]: failed to open stream: operation failed in...

How can I solve this…

Answer 1

From PHP’s doc

$zip = new ZipArchive;
if ($zip->open('test.zip') === TRUE) {
    $zip->extractTo($your_desired_dir);
    $zip->close();
    foreach (glob($your_desired_dir . DIRECTORY_SEPARATOR . 'New Folder') as $file) {
        $finfo = pathinfo($file);
        rename($file, $your_desired_dir . DIRECTORY_SEPARATOR . $finfo['basename']);
    }
    unlink($your_desired_dir . DIRECTORY_SEPARATOR . 'New Folder');
    echo 'ok';
} else {
    echo 'failed';
}

Dunno why are you using stream.