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存档文件中下一项

说明

zip_read(resource $zip): resource

读取ZIP存档文件中下一项。

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。

参见

zip_open() - 打开ZIP存档文件
zip_close() - 关闭一个ZIP档案文件
zip_entry_open() - 打开用于读取的目录实体
zip_entry_read() - 读取一个打开了的压缩目录实体

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User Contributed Notes 3 notes

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Anonymous ¶

2 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/

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Christian ¶

9 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.

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-20

nico at nicoswd dot com ¶

14 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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