Python 从请求获取url
以下是我的URL.py:Python 从请求获取url,python,django,Python,Django,以下是我的URL.py: from django.conf.urls import include, url from django.contrib import admin from stories.views import check, post urlpatterns = [ url(r'^admin/', admin.site.urls), url('^.*/$', check), ] 下面是我的观点.py: from django.http import HttpR
from django.conf.urls import include, url
from django.contrib import admin
from stories.views import check, post
urlpatterns = [
url(r'^admin/', admin.site.urls),
url('^.*/$', check),
]
下面是我的观点.py:
from django.http import HttpResponse
from django.shortcuts import render
from datetime import datetime
from models import Story
from .forms import StoryForm
def check(request):
try:
existing_story = Story.objects.get(name=?URLRequested?)
except Story.DoesNotExist:
return HttpResponse(post(request))
HttpResponse(existing_story.text)
我想转换输入的重定向404的url,并将其用作新对象的名称。如何做到这一点?最简单、最常见的方法是在url中创建一个参数,尽管
*
是一个非常广泛的正则表达式
url('^(?P<parameter_name>.*)/$', check),
def check(request, parameter_name):
try:
existing_story = Story.objects.get(name=parameter_name)
url(“^(?P.*)/$”,选中),
def检查(请求、参数名称):
尝试:
现有的\u故事=故事.对象.获取(名称=参数\u名称)
更常见的是查找特定的匹配项,例如\w+
,以查找名称或类似内容的文本字符串
有关更多信息,请参阅