PHP 8.3.4 Released!

zip_read

(PHP 4 >= 4.1.0, PHP 5 >= 5.2.0, PHP 7, PHP 8, PECL zip >= 1.0.0)

zip_readLegge la prossima voce in un archivio file zip

Descrizione

zip_read(resource $zip): resource

Legge la prossima voce in un archivio file zip.

Elenco dei parametri

zip

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 più file da leggere, oppure un errore se si verifica un errore.

Vedere anche:

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User Contributed Notes 3 notes

up
7
Anonymous
4 years ago
*Here is a simple example*
<?php
$zp
= zip_open('file.zip');

while (
$file = zip_read($zp)) {
echo
zip_entry_name($file).PHP_EOL;
}
?>

The output will be something similar to:

myfile.txt
mydir/
up
1
Christian
10 years ago
Note: Only the first 65535 entries will be returned, even if your archive contains more entries. See https://bugs.php.net/bug.php?id=59118 for details.
up
-17
nico at nicoswd dot com
16 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.

<?php

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

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

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

?>

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