Django学习-10-命名空间

时间：2021-08-20 22:14:14
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
 urlpatterns = [

     url(r'aaa/$', views.app01_aaa, name="index"),

 ]
 
app02_urls.py
 urlpatterns = [

     url(r'aaa/$', views.app02_aaa, name="index"),

 ]

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

 #app01

 /app01aaa/

 #app02

 /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

 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
 #http://127.0.0.1/aaa/aaa/

 # 结果：

 # /aaaaaa/

 #http://127.0.0.1/bbb/aaa/

 # 结果：

 # /bbbaaa/

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