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Python 通过URL传递参数时出现Django TypeError_Python_Django_Python 2.7_Django 1.3 - Fatal编程技术网

Python 通过URL传递参数时出现Django TypeError

Python 通过URL传递参数时出现Django TypeError,python,django,python-2.7,django-1.3,Python,Django,Python 2.7,Django 1.3,您好,我在Django(最新版本)中遇到以下错误: 当我按下主页上的链接以查看帖子时会发生这种情况。我试图传递帖子的id(我使用的是默认的[隐藏]主键,而不是我自己的自定义主键) 这是我的URL.py在索引和帖子页面中的外观: from django.conf.urls.defaults import patterns, include, url from journal.models import Post # Uncomment the next two lines to enable

您好,我在Django(最新版本)中遇到以下错误:

当我按下主页上的链接以查看帖子时会发生这种情况。我试图传递帖子的id(我使用的是默认的[隐藏]主键,而不是我自己的自定义主键)


这是我的URL.py在索引和帖子页面中的外观:

from django.conf.urls.defaults import patterns, include, url
from journal.models import Post


# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('journal.views',
    (r'^$', 'index'),
    (r'^post/(?P<id>\d+)/$', 'post'),
改变


我怀疑它会更好一点

啊,是的,我以前将其作为“id”使用,但是我一直缺少%s该部分的s,谢谢。看起来您最后忘记了%s?在你的原始版本中,你忘记了它,所以当你尝试上面的方法时,也许你没有改变它?
from django.conf.urls.defaults import patterns, include, url
from journal.models import Post


# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('journal.views',
    (r'^$', 'index'),
    (r'^post/(?P<id>\d+)/$', 'post'),
from django.http import HttpResponse
from journal.models import Post
from django.template import Context, loader
import os

# Hardcoded Varibles
SITE_ROOT = os.path.join(os.path.dirname(__file__))

# Create your views here.
def index(request):
    latest_post_list = Post.objects.all().order_by('-pub_date')[:10]
    t = loader.get_template(os.path.join(SITE_ROOT, 'templates', 'index.html'))
    c = Context({
    'latest_post_list': latest_post_list,
    })
    return HttpResponse(t.render(c))

def post(request, id):
    return HttpResponse("Hello this is post %" %(post_id))
def post(request, id):
    return HttpResponse("Hello this is post %" %(post_id))
def post(request, id):
    return HttpResponse("Hello this is post %s" % id)