PHP 中的压缩流

Zip Stream in PHP

提问by user371254

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?

我有一个 PHP 脚本，可以动态创建一个 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？

Thanks in advance.

提前致谢。

回答by weasel5i2

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.

如果您的 Web 服务器正在运行 linux，那么您可以在不生成临时文件的情况下进行流式传输。在 win32 下，您可能需要使用 Cygwin 或类似的东西。

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.

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 不要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。

<?
     $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 );
     }
?>

回答by Andreas Gohr

This library seems to be what you're looking for: https://github.com/maennchen/ZipStream-PHP

这个库似乎是你要找的：https: //github.com/maennchen/ZipStream-PHP

回答by Charles

If you're using the zip extension, the answer seems to be "no." The close method in the ZipArchive classis what triggers a write, and it seems to want to write to a file. You mighthave some luck using Streamsto simulate a file, but then you're keeping the entire thing in memory, and you still couldn't send it until you're done adding files.

如果您使用zip 扩展名，答案似乎是否定的。ZipArchive 类中的close 方法是触发写入的方法，它似乎想要写入文件。您可能会使用Streams来模拟一个文件，但随后您将整个内容保存在内存中，并且在您完成添加文件之前仍然无法发送它。

If you're having problems with timeouts or other problems while having a user wait for the zip file to be created, try a multi-step download process:

如果您在让用户等待创建 zip 文件时遇到超时或其他问题，请尝试多步骤下载过程：

1. User picks whatever they pick to create the zip
2. They're taken to a status page that uses ajax calls to actually build the zip file in the background. A pretty, distracting animation should keep their attention.
3. After the background processes build the zip, the user is redirected to a script that performs the download / redirected to the file.

1. 用户选择他们选择的任何东西来创建 zip
2. 他们被带到一个状态页面，该页面使用 ajax 调用在后台实际构建 zip 文件。一个漂亮的、分散注意力的动画应该吸引他们的注意力。
3. 在后台进程构建 zip 后，用户被重定向到执行下载/重定向到文件的脚本。

回答by Artur Grigio

Yes, you can directly stream the zip to the client using this git rep: Zip Stream Large Files

是的，您可以使用此 git rep 将 zip 直接流式传输到客户端：Zip Stream Large Files

回答by Artefacto

No, you have to use a temporary file, as in here.

不，你必须使用一个临时文件，在这里。