为什么这在编写 ValueError 时不起作用（尽管我使用了 Tango_with_django book 1.9）？

Question

我在 Django 1.9 教程中使用 Django 2.0.2，但在重建我的 URL 后无法弄清楚问题出在哪里。错误显示（视图rango.views.category 没有返回一个HttpResponse 对象，而是返回了None。）

from django.template import RequestContext
from django.shortcuts import render_to_response
from rango.models import Category, Page


def index(request):
    # Obtain the context from the HTTP request.
    context = RequestContext(request)

    # Query the database for a list of ALL categories currently stored.
    # Order the categories by no. likes in descending order.
    # Retrieve the top 5 only - or all if less than 5.

    # Place the list in our context_dict dictionary which will be passed to the template engine.
    # Query for categories - add the list to our context dictionary.
    category_list = Category.objects.order_by('-likes')[:5]
    context_dict = {'categories': category_list}

    # looping through each category returned, and create a URL attribute.
    # This attribute stores an encoded URL (e.g spaces replaced with underscores).
    for category in category_list:
        category.url = category.name.replace(' ', '_')

    # Render the response and return to the client!
    return render_to_response('rango/index.html', context_dict, context)


def category(request, category_name_url):
    # Request our context from the request passed to us.
    context = RequestContext(request)
    # Change underscores in the category name to spaces.
    # URLs don't handle spaces well, so we encode them as underscores.
    # We can then simply replace the underscores with spaces again to get the name.
    category_name = category_name_url.replace('_', ' ')
    # Create a context dictionary which we can pass to the template rendering engine.
    # We start by containing the name of the category passed by the user.
    context_dict = {'category_name': category_name}

    try:
        # Can we find a category with the given name?
        #  If we can't, the .get() method raises a DoesNotExist exception.
        # So the .get() method returns one model instance or raises an exception.
        category = Category.objects.get(name=category_name)
        # Retrieve all of the associated pages.
        #  Note that filter returns >= 1 model instance.
        pages = Page.objects.filter(category=category)
        # Adds our results list to the template context under name pages.
        context_dict['pages'] = pages
        # We also add the category object from the database to the context dictionary.
        # We'll use this in the template to verify that the category exists.
        context_dict['category'] = category

    except Category.DoesNotExist:
        # We get here if we didn't find the specified category.
        # Don't do anything - the template displays the "no category" message for us.
        pass
        # Go render the response and return it to the client.
        return render_to_response('rango/category.html', context_dict, context)

urls.py:

from django.urls import path, re_path
from . import views

urlpatterns = [
    path('', views.index, name='index'),
    re_path(r'^category/(?P<category_name_url>\w+)/$', views.category, name='category'),
]

Answer 1

更加相信错误消息。它说您的 category() 函数没有返回值，因为它没有返回值。它的return 缩进到与pass 相同的级别，因此它属于异常处理程序。将其缩进到 except 关键字的级别。