zip_read()

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

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

*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/

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.

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.