PHP 5.6.16 is available


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

zip_readLeer la siguiente entrada en el fichero ZIP


resource zip_read ( resource $zip )

Lee la siguiente entrada en un fichero zip.



Un fichero ZIP previamente abierto con zip_open().

Valores devueltos

Devuelve un recurso de una entrada de directorio para poder usar luego las funciones zip_entry_..., o FALSE si no hay más entradas a leer, o código de error en caso de ocurrir otro error.

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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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