1

在 CakePHP 控制器中,我有生成 ZIP 的逻辑,如下所示并且工作正常,我可以在服务器上找到 ZIP 存档:

function getAttachmentsInZip( $meetingId = null ){
    $this->autoRender = false;
    $this->request->allowMethod( ['post'] );
    $allTasksWithFiles_query = $this->Tasks->find('all')
        ->where(['Tasks.uploaded_file_path != ' => 'NULL']);
    $allTasksWithFiles = $allTasksWithFiles_query->toArray();
    $files = array();
    foreach( $allTasksWithFiles as $taskWithFile ){
        $files[] = $taskWithFile['uploaded_file_path'];
    }

    if( !empty( $files ) ){
        $destination = 'uploads/archives/meeting_' . $meetingId .'.zip';
        $zip = new ZipArchive();
        $zip->open( $destination, ZIPARCHIVE::CREATE | ZIPARCHIVE::OVERWRITE );

        foreach( $files as $file ){
            $zip->addFile('uploads/uploaded_files/' . $file, $file);
        }
        $zip->close();
    }
}

但是,我完全无法将存档直接返回到用户的浏览器。我得到的最接近的是以下片段,但存档已损坏且无法打开:

header("Content-type: application/zip");
header('Content-Disposition: attachment; filename="' . $destination . '"');
header("Pragma: no-cache");
header("Expires: 0");
readfile("asd");

非常感谢任何帮助或指导。如果我应该/可以提供更多代码 - 请询问。

4

0 回答 0