Question

我想在跨域中将带有jquery-file-upload（blueimp）的文件上传到symfony2应用程序。

为此，客户方：

SLjQuery(function () {
    'use strict';

    SLjQuery('#media_answer_answer').fileupload({
        url: "http://recrutonline.dev/app_dev.php/api/media/questionnaire-test-media/uploads",
        dataType: 'text',
        formData: SLjQuery('form#answer_process_form').serializeArray(),
        forceIframeTransport: true,
        redirect: 'http://mywebsite.dev/result.html?%s',
        done: function (e, data) {
            console.log('upload ok ', data)
        },
        progressall: function (e, data) {
            console.log(data.loaded/data.total);
            var progress = parseInt(data.loaded / data.total * 100, 10);
            SLjQuery('#progress .progress-bar').css(
                'width',
                progress + '%'
            );
        }
    }).prop('disabled', !SLjQuery.support.fileInput)
      .parent().addClass(SLjQuery.support.fileInput ? undefined : 'disabled')
    ;
});

在我的控制器symfony中：

public function postMediaUploadAction(Request $request)
{
    $requestData = $request->request->all();
    $redirectResponse = $requestData['redirect']; // http://mywebsite.dev/result.html?%s
    //...
    //here process on data & get file uploaded
    //...

    //example of data I would send back to the client : http://mywebsite.dev/result.html?{'file':[{'name':'filetoupload.jpg'}]}
    $response = str_replace("%s", "{'file':[{'name':'filetoupload.jpg'}]}", $redirectresponse); // is it the good method ?
    return ??
}

现在收到客户方回复：

在＆＃39;完成＆＃39;的console.log中ajax函数的参数：

data.result -> undefined
data.textStatus -> "success"

我试图返回很多东西，但结果仍未定义。所以你知道我必须返回什么样的数据吗？

Answer 1

您使用的是OneupUploaderBundle吗？

如果是这样，请不要将文件发布到您自己的控制器，而是将其发送到Oneup uploader endpoint。然后，您可以按implementing an event listener返回自定义数据。

例如：

namespace Foo\BarBundle\EventListener;

use Oneup\UploaderBundle\Event\PostPersistEvent;

class UploadListener
{
    public function onUpload(PostPersistEvent $event)
    {
        if ($file = $event->getFile()) {
            $response = $event->getResponse();
            $response['file']['name'] = $file->getKey();
        }
    }
}

在services.yml（或xml文件）中注册此事件侦听器：

# Event listener to handle uploaded files
foo_bundle.upload_listener:
    class: Foo\BarBundle\EventListener\UploadListener
    tags:
        - { name: kernel.event_listener, event: oneup_uploader.post_persist.default_uploader, method: onUpload }

当做想要自己处理上传时，我不建议您返回Symfony\Component\HttpFoundation\Response甚至更好的Symfony\Component\HttpFoundation\JsonResponse对象。请参阅documentation。

Jquery文件上传，如何用symfony2返回响应？

1 个答案: