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 — Retorna o nome do arquivo
Descrição
zip_entry_name
( resource
$zip_entry
) : stringRetorna o nome do arquivo especificado.
Valor Retornado
O nome do diretório de entrada.
Veja Também
- zip_open() - Abre um arquivo ZIPado
- zip_read() - Lê a próxima entrada em um arquivo ZIPado
add a note
User Contributed Notes 2 notes
leandro_dealmeida at hotmail dot com ¶
21 years ago
kevyn at opsone dot net ¶
15 years ago
Big note for filename with accents.
Some Zip softwares encode accents with CP850.
So use iconv for keeping your accents SAFE !
