问题描述

我有一个 PHP 脚本，可以动态创建 zip 文件并强制浏览器下载 zip 文件.问题是:我可以直接将 zip 文件写入连接到用户浏览器的输出流，而不是先将其保存为服务器上的真实文件，然后再发送文件吗?

I have a PHP script that creates a zip file on the fly and forces the browser to download the zip file. The question is: could I directly write the zip file to an output stream which is connected to the user's browser rather than save it as a real file on the server first and then send the file?

提前致谢.

推荐答案

如果您的 Web 服务器运行的是 linux，那么您可以在不生成临时文件的情况下进行流式处理.在 win32 下，您可能需要使用 Cygwin 或类似的东西.

If your web server is running linux, then you CAN do it streaming without a temp file being generated. Under win32 you may need to use Cygwin or something similar.

如果您使用-"作为 zip 文件名，它将压缩为标准输出.从那里，很容易使用 popen() 重定向该流.-q"参数只是告诉 zip 不输出它通常会输出的状态文本.有关详细信息，请参阅 zip(1) 联机帮助页.

If you use "-" as the zip file name, it will compress to stdout. From there, it's easy enough to redirect that stream using popen(). The "-q" argument simply tells zip to not output the status text it normally would. See the zip(1) manpage for more info.

<?
     $zipfilename = "zip_file_name.zip";

     if( isset( $files ) ) unset( $files );

     $target = "/some/directory/of/files/you/want/to/zip";

     $d = dir( $target );

     while( false !== ( $entry = $d->read() ) )
     {
       if( substr( $entry, 0, 1 ) != "." && !is_dir( $entry ) ) 
       {
         $files[] = $entry;
       }
     }

     header( "Content-Type: application/x-zip" );
     header( "Content-Disposition: attachment; filename="$zipfilename"" );

     $filespec = "";

     foreach( $files as $entry )
     {
       $filespec .= ""$entry" ";
     }

     chdir( $target );

     $stream = popen( "/usr/bin/zip -q - $filespec", "r" );

     if( $stream )
     {
       fpassthru( $stream );
       fclose( $stream );
     }
?>

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