PHP 5.4.31 Released

zip_entry_name

(PHP 4 >= 4.1.0, PHP 5 >= 5.2.0, PECL zip >= 1.0.0)

zip_entry_nameOttiene il nome della voce di una directory

Descrizione

string zip_entry_name ( resource $zip_entry )

Restituisce il nome della specifica voce della directory.

Elenco dei parametri

zip_entry

Una voce directory restituita da zip_read().

Valori restituiti

Il nome della voce della directory.

Vedere anche:

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User Contributed Notes 2 notes

up
0
kevyn at opsone dot net
5 years ago
Big note for filename with accents.

Some Zip softwares encode accents with CP850.

So use iconv for keeping your accents SAFE !
up
0
leandro_dealmeida at hotmail dot com
11 years ago
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
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