[py]django实现url跳转

HttpResponseRedirect实现uri跳转

需求: 访问/index 跳转到 /home

1.写url

from learn import viewsurlpatterns = [    url(r'^admin/', admin.site.urls),    url(r'^index/', views.index),    url(r'^home/', views.home),]

2.写views

from django.shortcuts import render,HttpResponseRedirect,HttpResponse# Create your views here.def index(request):    return HttpResponseRedirect("/home")def home(request):    return HttpResponse("home page")

3.测试访问

http://127.0.0.1:8000/index/ 跳转到了http://127.0.0.1:8000/home/

注: 跳转uri同时可以传参

return HttpResponseRedirect('/commons/invoice_return/index/?message=error')  #跳转到index界面

另官网说

The first argument to the constructor is required – the path to redirect to. This can be a fully qualified URL (e.g.’http://www.yahoo.com/search/‘) or an absolute path with no domain (e.g. ‘/search/’)。 参数既可以使用完整的url，也可以是绝对路径。

即

return HttpResponseRedirect("https://www.baidu.com/")  #访问http://127.0.0.1:8000/index/  跳转到了 https://www.baidu.com/

redirect+reverse重定向

需求: 访问/index 跳转到 /home

1.写urls

from learn import viewsurlpatterns = [    url(r'^admin/', admin.site.urls),    url(r'^index/', views.index),    url(r'^home/', views.home,name="home"),  # reverse是根据name字段解析的.]

2,写views

from django.shortcuts import render,HttpResponseRedirect,HttpResponsefrom django.core.urlresolvers import reversefrom django.shortcuts import redirect# Create your views here.def index(request):    # return HttpResponseRedirect("/home")    # return HttpResponseRedirect("https://www.baidu.com/")    return redirect(reverse('home', args=[]))  def home(request):    return HttpResponse("home page")

可以对比下HttpResponseRedirect实现方式:

def old_add2_redirect(request, a, b):    return HttpResponseRedirect(        reverse('add2', args=(a, b))    )