设计url 通过分发的方式 Xadmin_demo

Posted 2021-01-03 liang-jc

如

urlpatterns = [
    url(r‘^Xadmin/‘,([
　　　　　　　　　　url(r‘^add/$‘, views.add)
　　　　　　　　　　url(r‘^delete/$‘, views.delete)
　　　　　　　　　　], None,None)),
这样就完成了一级分发

然后进行自行设计url
需求 3张表，增删改查, 12个url
如
127.0.0.1:8000/Xadmin/app01/add/

127.0.0.1:8000/Xadmin/app01/delete/

在urls.py编辑

def list_view(request):
    return HttpResponse(‘userindex‘)
def add_view(request):
    return HttpResponse(‘user_add‘)
def change_view(request, id):
    return HttpResponse(‘user_change‘)
def delete_view(request, id):
    return HttpResponse(‘user_delete‘)

def has_urls():
    temp = []
    temp.append(url(r‘^$‘,list_view))
    temp.append(url(r‘^add/$‘,add_view))
    temp.append(url(r‘^(d+)/change/$‘,change_view))
    temp.append(url(r‘^(d+)/delete/$‘,delete_view))
    print(temp)
    return temp

def get_urls():
    print(admin.site._registry)
    temp = []
    for model,admin_class_obj in admin.site._registry.items():
        app_name = model._meta.app_label
        model_name = model._meta.model_name
        print(app_name,model_name)
        temp.append(url(r‘^{0}/{1}/‘.format(app_name,model_name),(has_urls(), None,None)),)
    return temp

urlpatterns = [
    path(‘admin/‘, admin.site.urls),
    url(r‘^Xadmin/‘,(get_urls(), None,None)),]

settings.py
设置
INSTALLED_APPS = [
‘app01.apps.App01Config‘,

‘app02.apps.App02Config‘,

]

app01项目下的admin.py

from django.contrib import admin
from app01 import models
# Register your models here.
admin.site.register(models.Book)
admin.site.register(models.Publish)

app01项目下的models.py

app02项目下的admin.py

from django.contrib import admin
from app02 import models
# Register your models here.
admin.site.register(models.Order)

app01项目下的models.py

from django.db import models
class Order(models.Model):
    title = models.CharField(max_length=32)
    def __str__(self):
        return self.title