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

zip_readLegge la prossima voce in un archivio file zip


resource zip_read ( resource $zip )

Legge la prossima voce in un archivio file zip.

Elenco dei parametri


Un file ZIP precedentemente aperto con zip_open().

Valori restituiti

Restituisce un puntatore alla risorsa directory da usarsi successivamente con le varie funzioni zip_entry_... oppure FALSE se non vi sono piĆ¹ file da leggere, oppure un errore se si verifica un errore.

Vedere anche:

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User Contributed Notes 2 notes

4 years ago
Note: Only the first 65535 entries will be returned, even if your archive contains more entries. See for details.
nico at nicoswd dot com
10 years ago
If you get an error like this:

Warning: zip_read() expects parameter 1 to be resource, integer given in xxxxxx on line x

It's because zip_open() failed to open the file and returned an error code instead of a resource. It took me a while to figure out why it failed to open the file, until I tried to use the FULL path to the file.


// Even if the file exists, zip_open() will return an error code.
$file = '';
$zip = zip_open($file);

// The workaround:
$file = getcwd() . '/';

// Or:
$file = 'C:\\path\\to\\';


This worked for me on Windows at least. I'm not sure about other platforms.
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