需求
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url = "https:// xxxxx.com/img/?img=https://vthumb.ykimg.com/0542040858B6D116000001813A0C32CD" |
如何实现呢？
urls 设置
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# django urls.py """cms_v1 URL Configuration The `urlpatterns` list routes URLs to views. For more information please see: https://docs.djangoproject.com/en/1.11/topics/http/urls/ Examples: Function views 1. Add an import: from my_app import views 2. Add a URL to urlpatterns: url(r'^$', views.home, name='home') Class-based views 1. Add an import: from other_app.views import Home 2. Add a URL to urlpatterns: url(r'^$', Home.as_view(), name='home') Including another URLconf 1. Import the include() function: from django.conf.urls import url, include 2. Add a URL to urlpatterns: url(r'^blog/', include('blog.urls')) """ from django.conf.urls import url,include from django.urls import path from .views import video_index,tag_index,video_details,search,cate_index,img_url urlpatterns = [ url(r'^img/$', img_url, name='img_url'), ] |
views.py 设置
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def img_url(request): img = request.GET.get('img') r = requests.get(img) return HttpResponse(r.content, content_type="image/jpeg") |
指定的 返回值要是二进制，content_type 指定要是 image/jpeg
支付宝扫一扫
:sho
学会了，支持 。