PythonDjango 支持 restful 风格 url

Posted 2020-08-15

URL通配符示例：

url(r‘^file_download/(?P<filename>(.)*)$‘, views.FILE_DOWNLOAD_VIEW.as_view()),

 

代码示例：

    def get(self, request, filename):
        from common.s3_storage import S3_STORAGE
        s3_storage = S3_STORAGE()
        key = s3_storage.package_bucket.get_key(filename)
        assert(key is not None)
        
        import tempfile
        temp = tempfile.NamedTemporaryFile()
        try:
            #temp = tempfile.TemporaryFile()
            key.get_contents_to_filename(temp.name)
            temp.seek(0)
            c = temp.read()
            
            #c = key.get_contents_as_string()
            #return APIResponse(status=status_code.success, data=OS_AND_KERNEL_MAPS)
            from django.http import StreamingHttpResponse
            #response = HttpResponse(c)
            response = StreamingHttpResponse(c)
            #response = StreamingHttpResponse( self.file_iterator(temp) )
            
            response[‘Content-Type‘] = ‘application/octet-stream‘
            import os
            response[‘Content-Length‘] = os.path.getsize(temp.name)
            response[‘Content-Disposition‘] = ‘attachment;filename="{0}"‘.format(filename)
            return response
        finally:
            temp.close()