django休息框架返回文件 [英] django rest framework return file

问题描述

我在我的views.py中有以下视图 -

  class FilterView（generics.ListAPIView）：
 
 vendor = self.request.GET.get（'vendor'）
文件名= self.request（自我，请求，格式=无） 。$。$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ 
 $ .get（'filedate'）
 snippets = model.objects.using（'markitdb'）。filter（Date__contains = filedate）
 serializer = cdx_compositesSerializer（snippets，many = True）
如果格式=='raw'：
 zip_file = open（'C：\temp\core\files\CDX_COMPOSITES_20140626.zip'，'rb'）
 response = HttpResponse（zip_file，content_type =' application / force-download'）
 response ['Content-Disposition'] ='attachment;文件名=％s'％'CDX_COMPOSITES_20140626.zip'
返回响应
 
其他：
返回响应（serializer.data）
  

它适用于xml，json，csv，但是当我尝试使用raw时，它不会返回文件，而是提供detail：Not发现为什么会发生这种情况？

我打的网址如下 -

工作 -

http：// dt-rpittom：8000 / testfilter /？vendor = markit& filename = cdx_composites.csv& filedate = 2014 -06-26& format = json

这应该返回一个zip文件进行下载。

http：// dt-rpittom：8000 / testfilter /？vendor = markit& filename = cdx_composites.csv& filedate = 2014-06-26& format = raw

解决方案

我不知道为什么我必须这样做 - 可能是Django Rest Framework内部的东西，允许定制我方式到格式？

我只是将其更改为以下 -

 如果fileformat =='raw'：
 zip_file = open（'C：\temp\core\files\CDX_COMPOSITES_20140626.zip'，'rb'）
 response = HttpResponse（FileWrapper zip_file），content_type ='application / zip'）
 response ['Content-Disposition'] ='attachment;文件名=％s'％'CDX_COMPOSITES_20140626.zip'
返回响应
  

然后在我的URL只是用新的值，它的工作正常。我很想知道为什么我不能使用格式来提供文件。

I have the following view in my views.py -

class FilterView(generics.ListAPIView):
    model = cdx_composites_csv

    def get(self, request, format=None):
        vendor = self.request.GET.get('vendor')
        filename = self.request.GET.get('filename')
        tablename = filename.replace(".","_")
        model = get_model(vendor, tablename)
        filedate = self.request.GET.get('filedate')        
        snippets = model.objects.using('markitdb').filter(Date__contains=filedate)
        serializer = cdx_compositesSerializer(snippets, many=True)
        if format == 'raw':
            zip_file = open('C:\temp\core\files\CDX_COMPOSITES_20140626.zip', 'rb')
            response = HttpResponse(zip_file, content_type='application/force-download')
            response['Content-Disposition'] = 'attachment; filename="%s"' % 'CDX_COMPOSITES_20140626.zip'
            return response

        else:
            return Response(serializer.data)

It works great for xml, json, csv but when I try to use raw it's not returning the file instead it's giving ""detail": "Not found"" why is this happening?

The URL I'm hitting is as follows -

json example that works -

http://dt-rpittom:8000/testfilter/?vendor=markit&filename=cdx_composites.csv&filedate=2014-06-26&format=json

This should return a zip file for download.

http://dt-rpittom:8000/testfilter/?vendor=markit&filename=cdx_composites.csv&filedate=2014-06-26&format=raw

解决方案

I don't know why I had to do this - may be something internal to Django Rest Framework that doesn't allow putting custom methods onto format?

I simply changed it to the following -

if fileformat == 'raw':
    zip_file = open('C:\temp\core\files\CDX_COMPOSITES_20140626.zip', 'rb')
    response = HttpResponse(FileWrapper(zip_file), content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename="%s"' % 'CDX_COMPOSITES_20140626.zip'
    return response

Then in my URL just hit with the new value and it works fine. I'd love to know why I can't use format though to serve a file.

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