The 7th Annual China PHP Conference


(PHP 4 >= 4.1.0, PHP 5 >= 5.2.0, PHP 7, PECL zip >= 1.0.0)

zip_readLeer la siguiente entrada en el fichero ZIP


zip_read ( resource $zip ) : resource

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

6 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
11 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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