Android-Django图片上传

问题描述 投票:9回答:2

我试图上传图像与android作为前端和django作为后端。

该模型:

    class Photo(models.Model):
        title = models.CharField(max_length=255,blank=True)
        photo = models.FileField(upload_to='photos')
        description = models.TextField(blank=True)
        uploaded = models.DateTimeField(auto_now_add=True)
        modified = models.DateTimeField(auto_now=True)

        class Meta:
            db_table = 'media_photos'

        def __unicode__(self):
            return '%s' % self.title

url url的视图(r'^ photos / upload / $','upload_photo'):

def upload_photo(request):
form=PhotoForm(request.POST,request.FILES)
if request.method=='POST':
    if form.is_valid():
        image = request.FILES['photo']
        title1 =''
        new_image = Photo(title=title1,photo=image,description='')
        new_image.save()
        response_data=[{"success": "1"}]
        return HttpResponse(simplejson.dumps(response_data), mimetype='application/json')

现在我想从android访问这里的视图。所以现在我上传图像的android端代码是:

     public void doFileUpload(String path){
        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        DataInputStream inStream = null;
        String lineEnd = "\r\n";
        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;
        int maxBufferSize = 1*1024*1024;
        String urlString = "http://";   // server ip
        try
        {
         //------------------ CLIENT REQUEST
        FileInputStream fileInputStream = new FileInputStream(new File(path) );
         // open a URL connection to the Servlet
         URL url = new URL(urlString);
         // Open a HTTP connection to the URL
         conn = (HttpURLConnection) url.openConnection();
         // Allow Inputs
         conn.setDoInput(true);
         // Allow Outputs
         conn.setDoOutput(true);
         // Don't use a cached copy.
         conn.setUseCaches(false);
         // Use a post method.
         conn.setRequestMethod("POST");
         conn.setRequestProperty("Connection", "Keep-Alive");
         conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+"    ");
         dos = new DataOutputStream( conn.getOutputStream() );
         dos.writeBytes(lineEnd);
         dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + path + "\"" + lineEnd);
         dos.writeBytes(lineEnd);

         // create a buffer of maximum size
         bytesAvailable = fileInputStream.available();
         bufferSize = Math.min(bytesAvailable, maxBufferSize);
         buffer = new byte[bufferSize];

         // read file and write it into form...
         bytesRead = fileInputStream.read(buffer, 0, bufferSize);
         while (bytesRead > 0)
         {
          dos.write(buffer, 0, bufferSize);
          bytesAvailable = fileInputStream.available();
          bufferSize = Math.min(bytesAvailable, maxBufferSize);
          bytesRead = fileInputStream.read(buffer, 0, bufferSize);
         }

         // send multipart form data necesssary after file data...
         dos.writeBytes(lineEnd);
         dos.writeBytes(lineEnd);

         // close streams
         Log.e("Debug","File is written");
         fileInputStream.close();
         dos.flush();
         dos.close();
        }
        catch (MalformedURLException ex)
        {
             Log.e("Debug", "error: " + ex.getMessage(), ex);
        }
        catch (IOException ioe)
        {
             Log.e("Debug", "error: " + ioe.getMessage(), ioe);
        }

        //------------------ read the SERVER RESPONSE
        try {
              inStream = new DataInputStream ( conn.getInputStream() );
              String str;
              while (( str = inStream.readLine()) != null)
              {
                   Log.e("Debug","Server Response "+str);
              }
              inStream.close();
        }
        catch (IOException ioex){
             Log.e("Debug", "error: " + ioex.getMessage(), ioex);
        }
      }
}

但它给了我一个错误:

E/Debug(590): error: java.net.URISyntaxException: Authority expected at index 7: http://
android django file-upload
2个回答
2
投票

应该是urlString = "http://192.168.1.2/photos/upload";

但是如果它不起作用,你会得到一个不同的错误,我们可能需要该错误才能进一步回答。

此外,您似乎没有真正的边界字符串集,并且您没有正确使用它。

http://stunningco.de/2010/04/25/uploading-files-to-http-server-using-post-android-sdk/,注意他如何使用唯一的边界字符串,并将其写入输出流?

您应该开始将问题标记为已回答。当你这样做时,你会得到更好的成功。


0
投票

即使你克服了这些困难,你也会违背Django的要求,即你提供一个带有POST参数的csrfmiddlewaretoken。我不知道你如何在Android设备上获得它;通过设计,令牌是为了防止从除Django前端之外的任何东西(即“模板”)调用Django后端代码(即“视图”)。即它旨在阻止你正在尝试做的事情。

您可以在特定视图上禁用csrf功能 - 使用“@csrf_exempt”装饰器。然后你可以决定你是否足够关心安全性,找出你自己的替代品来解决csrf给你的东西。

或者,不是从Android应用程序上传图片,而是编写一个Web应用程序来上传图片,并让您的Django项目提供该Web应用程序。您的Android应用程序可以启动浏览器(作为Intent)并将其指向该Web应用程序。这里有一些代码可以做到:https://simpleisbetterthancomplex.com/tutorial/2016/08/01/how-to-upload-files-with-django.html(它不会赢得任何选美比赛,但确实有效。)

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