文件上传使用Django [英] FileUpload with Django

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本文介绍了文件上传使用Django的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我使用的是 AJAX上传 code做了一个简单的AJAX文件上传。我来跨越的问题是,该文件没有显示在后台提交后。

前端code是pretty的基础:

 < D​​IV ID =image_uploader>上传更多图片< / DIV>
<脚本类型=文/ JavaScript的字符集=utf-8>
    功能createUploader(){
        VAR上传=新qq.FileUploader({
            元素:的document.getElementById('image_uploader),
            动作:/添加/图像/ 1',
            调试:真实,
            的onsubmit:函数(){
                progress.show();
            },
            的onComplete:函数(){
                progress.hide();
            },
            OnCancel的:函数(){
                progress.hide();
            },
        });
    };

    createUploader();
< / SCRIPT>
 

后端code(目前正在进行中)也pretty的基础:

 高清add_image(请求ID):
    打印请求
    如果request.FILES:
        返回的Htt presponse({成功:真})
    其他:
        返回的Htt presponse({成功:假消息:找不到FILES})
 

解决方案

对于我来说,使用code从的亚历克斯·库尔 request.GET中['qqfile'] 有文件名和 request.read()(在Django 1.3)返回的数据。

request.FILES只适用于尚未没有发生对我来说,一个场景。我使用AJAX上传直接对话Photologue,和我的code看起来是这样的:

 高清save_upload(上传,文件名,raw_data):
    
    raw_data:如果为真,upfile是生后数据的Htt prequest对象
    作为文件,而不是一个Django UploadedFile的从request.FILES
    
    尝试:
        文件名= os.path.normpath(os.path.join(IMAGE_UPLOAD_PATH,文件名))
        使用的BufferedWriter(FileIO专注(文件名,世行))为DEST:
            #如果先进的上传,直接从HTTP请求读取
            #与Django的1.3功能
            如果raw_data:
                (目录名,文件名)= os.path.split这样的(文件名)
                (fileBaseName,fileExtension)= os.path.splitext(文件名)
                #
                #在这里,如果fileBaseName小于n个字符,可能要就打上只是为了好玩日期
                #
                尝试:
                    i_can_has_p = Photo.objects.get(标题= fileBaseName)
                    标题= fileBaseName +_+ STR(datetime.datetime.now()的strftime(%Y%M%胸苷%H%M%S))
                除了Photo.DoesNotExist:
                    标题= fileBaseName
                title_slug = slugify(标题)
                P =照片(标题=标题,title_slug = title_slug)
                p.image.save(文件名,ContentFile(uploaded.read()))
            #如果不生,这是一个形式上传,以便读取普通的Django豆腐块时尚
            其他:
                #TODO:弄清楚这个时候被调用,使其工作保存成图片如上
                对于C在uploaded.chunks():
                    dest.write(三)
    除了IO错误:
        #无法最有可能打开文件
        返回False
    返回True

高清ajax_upload(要求):
  如果request.method ==POST:
      #AJAX上传将通过在查询字符串的文件名,如果它是高级AJAX上传
      如果request.is_ajax():
          #该文件被存储的原始的请求
          上传=请求
          is_raw = TRUE
          尝试:
              文件名= request.GET中['qqfile']
          除了KeyError异常:
              返回的Htt presponseBadRequest(AJAX请求无效)
      #不是一个ajax上传,于是就出现了基本的iframe版本通过表单提交
      其他:
          is_raw =假
          如果len(request.FILES)== 1:
              #FILES是在Django一本字典,而是阿贾克斯上传给了上传文件的
              #的ID的基础上随机数,所以它在这里不能猜测在code。
              #而不是编辑阿贾克斯上传来传递ID的查询字符串,请注意
              #每次上传的一个单独的请求,以便文件应该只有一个入口。
              #因此,我们可以只抢到第一(也是唯一一个)在字典的价值。
              上传= request.FILES.values​​()[0]
          其他:
              提高HTTP404(坏上传)
          文件名= upload.name

  #保存文件
  成功= save_upload(上传,文件名,is_raw)

  #让阿贾克斯上传知道我们是否保存与否
  ret_json = {成功:成功,}
  返回的Htt presponse(json.dumps(ret_json))
 

在我的情况下, ajax_upload 是称为AJAX的动作的功能:参数

I'm using the ajax-upload code to do a simple AJAX file upload. The issue I'm coming across is the file isn't showing up on the backend after submitting.

The frontend code is pretty basic:

<div id="image_uploader">Upload More Images</div>
<script type="text/javascript" charset="utf-8">
    function createUploader(){            
        var uploader = new qq.FileUploader({
            element: document.getElementById('image_uploader'),
            action: '/add/image/1',
            debug: true,
            onSubmit : function () {
                progress.show();
            },
            onComplete : function () {
                progress.hide();
            },
            onCancel : function () {
                progress.hide();
            },
        });           
    };

    createUploader();
</script>

The backend code (currently in progress) is also pretty basic:

def add_image(request, id):
    print request
    if request.FILES:
        return HttpResponse("{success:true}")
    else:
        return HttpResponse("{success:false, message:'Unable to find FILES}")

解决方案

For me, using code from Alex Kuhl, request.GET['qqfile'] had the filename and request.read() (in Django 1.3) returned the data.

request.FILES was only used in a scenario that hasn't yet happened for me. I'm using ajax-upload to talk directly to Photologue, and my code looks something like this:

def save_upload( uploaded, filename, raw_data ):
    """
    raw_data: if True, upfile is a HttpRequest object with raw post data
    as the file, rather than a Django UploadedFile from request.FILES
    """
    try:
        filename = os.path.normpath(os.path.join(IMAGE_UPLOAD_PATH, filename))
        with BufferedWriter( FileIO( filename, "wb" ) ) as dest:
            # if the "advanced" upload, read directly from the HTTP request
            # with the Django 1.3 functionality
            if raw_data:
                (dirName, fileName) = os.path.split(filename)
                (fileBaseName, fileExtension)=os.path.splitext(fileName)
                #
                # right here, if fileBaseName is less than n characters, might want to slap on a date just for fun
                #
                try:
                    i_can_has_p = Photo.objects.get(title=fileBaseName)
                    title = fileBaseName + "_" + str(datetime.datetime.now().strftime("%Y%m%dT%H%M%S"))
                except Photo.DoesNotExist:
                    title = fileBaseName
                title_slug = slugify(title)
                p = Photo(title=title, title_slug=title_slug)
                p.image.save(filename,ContentFile(uploaded.read()))
            # if not raw, it was a form upload so read in the normal Django chunks fashion
            else:
                # TODO: figure out when this gets called, make it work to save into a Photo like above
                for c in uploaded.chunks( ):
                    dest.write( c )
    except IOError:
        # could not open the file most likely
        return False
    return True

def ajax_upload( request ):
  if request.method == "POST":
      # AJAX Upload will pass the filename in the querystring if it is the "advanced" ajax upload
      if request.is_ajax( ):
          # the file is stored raw in the request
          upload = request
          is_raw = True
          try:
              filename = request.GET[ 'qqfile' ]
          except KeyError:
              return HttpResponseBadRequest( "AJAX request not valid" )
      # not an ajax upload, so it was the "basic" iframe version with submission via form
      else:
          is_raw = False
          if len( request.FILES ) == 1:
              # FILES is a dictionary in Django but Ajax Upload gives the uploaded file an
              # ID based on a random number, so it cannot be guessed here in the code.
              # Rather than editing Ajax Upload to pass the ID in the querystring, note that
              # each upload is a separate request so FILES should only have one entry.
              # Thus, we can just grab the first (and only) value in the dict.
              upload = request.FILES.values( )[ 0 ]
          else:
              raise Http404( "Bad Upload" )
          filename = upload.name

  # save the file
  success = save_upload( upload, filename, is_raw )

  # let Ajax Upload know whether we saved it or not
  ret_json = { 'success': success, }
  return HttpResponse( json.dumps( ret_json ) )

In my case, ajax_upload is the function called by ajax's action: parameter

这篇关于文件上传使用Django的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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