Question

您好我在Django（最新版本）中收到以下错误：

TypeError at /post/1/
post() got an unexpected keyword argument 'post_id'

当我按下主页上的链接以查看帖子的自我时，我会尝试传递帖子的ID（我使用默认的[隐藏]主键，而不是我自己的自定义键）< / p>

这就是我的urls.py对索引和帖子页面的看法：

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'),

这是我的views.py：

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))

Answer 1

更改

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

到

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

我怀疑它会更好一点！

通过URL传递参数时的Django TypeError

1 个答案: