使用fopen（）包装器创建ZIP文件

Question

如何使用fopen() wrapper创建ZIP文件？这显然是不的方式：

<?php

if( class_exists('ZipArchive') ){
    echo 'Class ZipArchive exists, generating file...' . PHP_EOL;

    $fp = fopen('zip://' . dirname(__FILE__) . '/test.zip', 'w');
    if($fp){
        fwrite($fp, 'Lorem ipsum dolor sit amet, consectetur adipisicing elit.');
        fclose($fp);
        echo 'Done' . PHP_EOL;
    }else{
        echo 'Could not open file' . PHP_EOL;
    }
}else{
    echo 'Class Zip is not available' . PHP_EOL;
}

...因为我得到的只是：

Class ZipArchive exists, generating file...

Warning: fopen(zip://C:\tmp/test.zip) [http://es.php.net/function.fopen]: failed to open stream: operation failed in C:\tmp\test.php on line 6
Could not open file

Answer 1

我终于认为zip:包装器支持写作的主张是文档错误和reported it as such。错误报告已被接受并修复。