Django中提供了一个关于URL的映射的解决方案，
1.客户端的浏览器发起一个url请求，Django根据URL解析，把url中的参数捕获，调用相应的试图，获取相应的数据，然后返回给客户端显示
2.通过一个视图的名字，再加上一些参数和值，逆向获取相应的URL
第一个就是平常的请求有URLconf来解析的过程，
第二个叫做，url的逆向解析，url逆向匹配，url的逆向查阅，等
Django提供了不同的层级的url逆向处理工具：
1.在模板templates中，使用url标记，如：{% url %}
2.在Python代码中，使用django.core.urlresolvers.reverse()方法
3.在更高一层级的处理url中，用get_absolute_url()方法
新一个项目diango
访问longin页面跳转到 index页面
app01 views.py
from django.shortcuts import render,redirect,HttpResponsedef index(request): ???return HttpResponse(‘Index‘)def login(request): ???return HttpResponse(‘longin‘)
urls.py
from django.conf.urls import urlfrom app01 import viewsurlpatterns = [ ???url(r‘^index/‘, views.index), ???url(r‘^login/‘, views.login),]
在平常做法
views.py
def login(request): ???return redirect(‘/index/‘)
如果usr 很长 urlpatterns = [ ???url(r‘^index/xxxxx/xxx/xxx/xxx/xx/‘, views.index), ???url(r‘^login/‘, views.login),]
可以用平常做法：
app01 views.py
def login(request): ???return redirect(‘/index/xxxxx/xxx/xxx/xxx/xx/‘)
这种方法看上去不是那么的友好, 我们用usl 别名的方式
导入reverse
app01/views.py
from django.urls import reversedef login(request): ???url = (‘inx‘) ???return redirect(url)
urls.py
urlpatterns = [ ???url(r‘^index/xxxxx/xxx/xxx/xxx/xx/‘, views.index,name=‘inx‘),]
通过 usl名称，反向生成usl
include方法来实现路由转发功能
urls.py
url(r‘^cmdb/‘, include’(crm.urls)’),
crm/urls.py
from django.conf.urls import url,includefrom django.contrib import adminfrom crm import viewsurlpatterns = [ ???url(r‘^hosts/‘, views.hosts,name=‘hosts‘),]
crm/views.py
def hosts(request): ??return HttpResponse(‘主机列表‘)
通过这个cmdb前缀分发，分发到自己的urls中，好处可以将各种的业务拆分
反向生成时，使用reverse(‘namespace:name’)
新建app, monitor 、openstack ,
url(r‘^monitor/‘, include(‘monitor.urls‘,namespace=‘m‘)),url(r‘^openstack/‘, include(‘openstack.urls‘,namespace=‘o‘)),
monitor/ulrs.py
from django.conf.urls import url,includefrom django.contrib import adminfrom monitor import viewsurlpatterns = [ ???url(r‘^hosts/‘, views.hosts,name=‘hosts‘),]
monitor/views.py
from django.shortcuts import render,HttpResponsedef hosts(request): ???return HttpResponse(‘监控系统，主机列表‘)
openstack /ulrs.py
from django.conf.urls import url,includefrom django.contrib import adminfrom openstack import viewsurlpatterns = [ ???url(r‘^hosts/‘, views.hosts,name=‘hosts‘),]
openstack /views.py
from django.shortcuts import render,HttpResponsedef hosts(request): ???return HttpResponse(‘OpenStack，主机列表‘)登录成功时跳转到 openstack/主机列表页面
def login(request): ???return HttpResponse(‘Login‘) ???# 跳转到 openstack主机列表页面 ???# return redirect(‘/openstack/hosts/‘) ???url = reverse(‘o:hosts‘)
include本质 【就是元组，一级一级做分发】
urls.py
from monitor import views as mviewsfrom openstack import views as oviews ???# /monitor/hosts/ ???url(r‘^login/‘, views.login), ???????url(r‘^monitor/‘, ([ ???????????????????????????????url(r‘^hosts/‘,mviews.hosts,name=‘hhhhh‘), ???????????????????????????????url(r‘^c1/‘,mviews.hosts), ???????????????????????????????url(r‘^x1/‘, ([ ???????????????????????????????????????????????url(r‘^xxx1/‘,mviews.hosts), ???????????????????????????????????????????????url(r‘^xxx2/‘,mviews.hosts,name=‘xx2‘), ???????????????????????????????????????????????url(r‘^xxx3/‘,mviews.hosts), ???????????????????????????????????????????????url(r‘^xxx4/‘,mviews.hosts), ???????????????????????????????????????????],None,None)), ???????????????????????????????url(r‘^c2/‘,mviews.hosts), ???????????????????????????????url(r‘^c3/‘,mviews.hosts), ???????????????????????????],None,‘mm‘)), ???????url(r‘^openstack/‘, ([ ???????????????????????????????url(r‘^hosts/‘,oviews.hosts,name=‘hhhhh‘), ???????????????????????????????url(r‘^c1/‘,oviews.hosts), ???????????????????????????????url(r‘^c2/‘,oviews.hosts), ???????????????????????????????url(r‘^c3/‘,oviews.hosts), ???????????????????????????],None,‘oo‘)),]
views.py
???# 跳转到 openstack主机列表页面 ????url = reverse(‘oo:hhhhh‘) ????url = reverse(‘mm:uu:xx2‘)
动态生成url
views.py
def login(request): ???return HttpResponse(‘Login‘)
urls.py
urlpatterns = [ ???url(r‘^login/‘, views.login),]
for i in range(10): ???temp = [] ???for j in range(5): ???????temp.append(url(r‘^inner_%s/‘ % j, views.login)) ???v = url(r‘^login_%s/‘ %i, (temp,None,None)) ???urlpatterns.append(v)
Django- 反向生成url
原文地址：https://www.cnblogs.com/qiangyuge/p/8306534.html