PHP 8.1.28 Released!

zip_read

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

zip_read读取 ZIP 文件归档中下一项

警告

本函数已自 PHP 8.0.0 起被废弃。强烈建议不要依赖本函数。

说明

zip_read(resource $zip): resource|false

读取 ZIP 文件归档中下一项。

参数

zip

一个ZIP压缩文件,该ZIP归档文件之前应由函数 zip_open() 打开。

返回值

成功的时候返回该当前实体资源供zip_entry_... 系列函数后续使用; 如果没有更多的读取项,则会返回 false 如果遇到错误则会返回相应的错误码。

更新日志

版本 说明
8.0.0 弃用此函数,取而代之的是对象 API,请参阅 ZipArchive::statIndex()

参见

add a note

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
11 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.
To Top