Django学习-10-命名空间

不使用命名空间，且两个APP某条url使用相同的name属性

app01_urls.py

1 urlpatterns = [
2     url(r'aaa/$', views.app01_aaa, name="index"),
3 ]

app02_urls.py

1 urlpatterns = [
2     url(r'aaa/$', views.app02_aaa, name="index"),
3 ]

各自的函数中同时反解URL

#app01.views
def aaa(request):
    return  HttpResponse(reverse("index"))

#app02.views
def aaa(request):
    return  HttpResponse(reverse("index"))

页面显示结果app01.views

1 #app01
2 
3 /app01aaa/
4 
5 
6 #app02
7 
8 /app01aaa/

总结

由于name没有作用域，Django在反解URL时，会在项目全局顺序搜索，当查找到第一个name指定URL时，立即返回

我们在开发项目时，会经常使用name属性反解出URL，当不小心定义相同的name时，可能会导致URL反解错误，为了避免这种事情发生，引入了命名空间

解决

使用Django提供namespace参数

使用命名空间

app.urls必须添加变量

1 app_name = 'app01'

project_urls.py

from django.conf.urls import url,include

urlpatterns = [
    url(r'^aaa', include("app01.urls"),namespace='aaa')),
    url(r'^bbb', include("app02.urls"),namespace='bbb')),
]

app01_urls.py

from django.conf.urls import url
from app01 import views

app_name = "app01"

urlpatterns = [
    url(r'aaa/$', views.aaa, name="index"),
]

app02_urls.py

from django.conf.urls import url
from app01 import views

app_name = "app02"

urlpatterns = [
    url(r'aaa/$', views.aaa, name="index"),
]

app01_views.py

def aaa(request):
    return HttpResponse(reverse("app01:index"))

app02_views.py

def aaa(request):
    return HttpResponse(reverse("app02:index"))

验证

 1 #http://127.0.0.1/aaa/aaa/
 2 
 3 # 结果：
 4 # /aaaaaa/
 5 
 6 
 7 #http://127.0.0.1/bbb/aaa/
 8 
 9 # 结果：
10 # /bbbaaa/