PHP 5.6.14 is available


(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

2 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
8 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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