我对drf有一些问题,有人可以指出我做错了什么吗?
这是我的视图函数,我也实现了easy_thumbnails,以便在提出此类请求时可以裁剪图像。
from __future__ import unicode_literals
from django.shortcuts import render
from rest_framework import viewsets
from rest_framework.response import Response
from .models import Image
from .serializers import ImageSerializer
from easy_thumbnails.files import get_thumbnailer
class ImageViewSet(viewsets.ModelViewSet):
queryset = Image.objects.all()
serializer_class = ImageSerializer
def retrieve(self, request, pk=None):
height = request.GET.get('height', None)
width = request.GET.get('width', None)
print("height = {}".format(height))
print("width = {}".format(width))
print("id = {}".format(pk))
img = Image.objects.get(pk = pk)
if height and width:
options = {'size': (height, width), 'crop': True}
thumb_url = get_thumbnailer(img.image).get_thumbnail(options).url
else:
thumb_url = get_thumbnailer(img.image).url
return Response(thumb_url)
现在,如果转到http://127.0.0.1:8000/api/images/
它会返回我一个图像列表
并且http://127.0.0.1:8000/api/images/1/?height=320&width=420
返回类似
HTTP 200 OK
Allow: GET, PUT, PATCH, DELETE, HEAD, OPTIONS
Content-Type: application/json
Vary: Accept
"/media/10438039923_2ef6f68348_c.jpg.320x420_q85_crop.jpg"
我想要这样的带有字段名称的响应。
{
"title": "Item 1",
"description": "Description 1",
"image": "http://127.0.0.1:8000/media/10438039923_2ef6f68348_c.jpg"
}
如何解决此问题?我是drf和django的新手
这是我的序列化程序类
from rest_framework import serializers
from .models import Image
class ImageSerializer(serializers.ModelSerializer):
class Meta:
model = Image
fields = [
'title',
'description',
'image',
]
使用这个:
def retrieve(self, request, pk=None):
height = request.query_params.get('height', None)
width = request.query_params.get('width', None)
img = self.get_object()
if height and width:
options = {'size': (height, width), 'crop': True}
thumb_url = get_thumbnailer(img.image).get_thumbnail(options).url
else:
thumb_url = get_thumbnailer(img.image).url
serializer = self.get_serializer(img)
# insert thumb_url into data if you need it
response_dict = {}
response_dict.update(serializer.data)
response_dict['image'] = thumb_url
return Response(response_dict)
你可以
这样做,
return Response({"title": img.title, "description": img.description,
"image": img.image, "thumbnail": thumb_url})