If you want to get the real name of the file without the directory name, you can just use the function basename() as the follow:
<?
$zip_dir = "./import/";
$zip = zip_open($zip_dir."import.zip");
if ($zip) {
while ($zip_entry = zip_read($zip)) {
$file = basename(zip_entry_name($zip_entry));
$fp = fopen($zip_dir.basename($file), "w+");
if (zip_entry_open($zip, $zip_entry, "r")) {
$buf = zip_entry_read($zip_entry, zip_entry_filesize($zip_entry));
zip_entry_close($zip_entry);
}
fwrite($fp, $buf);
fclose($fp);
echo "The file ".$file." was extracted to dir ".$zip_dir."\n<br>";
}
zip_close($zip);
}
?>
Thefore you can extract files without concern with the directory that is set inside the zip source.
Remember to give write permission (w) on that directory.
Hello from Brazil.
Leandro
zip_entry_name
(PHP 4 >= 4.1.0, PHP 5 >= 5.2.0, PHP 7, PHP 8, PECL zip >= 1.0.0)
zip_entry_name — Ottiene il nome della voce di una directory
Descrizione
Restituisce il nome della specifica voce della directory.
Valori restituiti
Il nome della voce della directory.
Vedere anche:
- zip_open() - Apre un archivio ZIP
- zip_read() - Legge la prossima voce in un archivio file zip
add a note
User Contributed Notes 2 notes
leandro_dealmeida at hotmail dot com ¶
22 years ago
kevyn at opsone dot net ¶
16 years ago
Big note for filename with accents.
Some Zip softwares encode accents with CP850.
So use iconv for keeping your accents SAFE !
