Django得到了一个意想不到的关键字参数'id'

Question

我正在尝试在Django中创建一个电话簿.我的urls.py:

    urlpatterns = [
    url(r'^$', views.people_list, name='people_list'),
    url(r'^(?P<id>\d)/$', views.person_detail, name='person_detail'),
]

views.py:

def people_list(request):
    people = Person.objects.all()
    return render(request, 'phonebook/person/list.html', {'people': people})


def person_detail(request, person):
    person = get_object_or_404(Person, id=person)
    return render(request, 'phonebook/person/detail.html', {'person': person})

来自models.py:

def get_absolute_url(self):
    return reverse('phonebook:person_detail', args=[self.id])

和list.html:

{% block content %}
<h1>Phonebook</h1>
{% for person in people %}
<h2>
    <a href="{{ person.get_absolute_url }}">
        {{ person.name }} {{ person.last_name }}
    </a>
</h2>
<p class="where">
{{ person.department }}, {{ person.facility }}
    </p>
{{ person.phone1 }}<br>
{% endfor %}
{% endblock %}

该索引看起来不错,但是当我尝试点击链接获取person_detail网站时,我收到以下消息:

/ phonebook/4/person_detail()中的TypeError获得了一个意外的关键字参数'id'请求方法:GET请求URL:http://127.0.0.1:8000/phonebook/4/ Django版本:1.9.6异常类型:TypeError异常值:person_detail()得到一个意外的关键字参数'id'

我在urls.py和函数get_absolute_url中有'id'参数.我不明白什么是错的.

Answer 1

您的参数?P<id>中的URL映射必须在视图中匹配的参数def person_detail(request, person):

他们应该是id或person.

Answer 2

您应该修复视图并使用id参数名称而不是person：

def person_detail(request, id):