Python list_of_post（）获取了意外的关键字参数'；slug'；

我正在学习一个教程，但是这个教程的django版本与我的2.0.5版本不同。所以，我得到了这个问题：

list_of_post() got an unexpected keyword argument 'slug'
Request Method: GET
Request URL:    http://127.0.0.1:8000/blog/a-test/
Django Version: 2.0.5
Exception Type: TypeError
Exception Value:    
list_of_post() got an unexpected keyword argument 'slug'
Exception Location: /Users/sumeixu/anaconda3/lib/python3.6/site-packages/django/core/handlers/base.py in _get_response, line 126
Python Executable:  /Users/sumeixu/anaconda3/bin/python
Python Version: 3.6.3
Python Path:    
['/Users/sumeixu/djangotest',
 '/Users/sumeixu/anaconda3/lib/python36.zip',
 '/Users/sumeixu/anaconda3/lib/python3.6',
 '/Users/sumeixu/anaconda3/lib/python3.6/lib-dynload',
 '/Users/sumeixu/anaconda3/lib/python3.6/site-packages',
 '/Users/sumeixu/anaconda3/lib/python3.6/site-packages/aeosa']
Server time:    Thu, 7 Jun 2018 15:00:29 +0000

blog/view.py

def post_detail(request,slug):
    post = get_object_or_404(Post,slug=slug)
    template = 'blog/post/post_detail.html'
    return render(request,template,{'post':post}) 

post_of_detail.html

{% extends 'blog/post/base.html' %}

{% block title %}{{post.seo_title}}{% endblock %}

{% block content %}
         <h2>{{ post.title }}</h2>
         <p>Written by {{post.author }} on {{post.published}}</p>
         <hr>
         {{posts.content}
{% endblock %}

{%extends'blog/post/base.html%}
{%block title%}{{post.seo_title}{%endblock%}
{%block content%}
{{post.title}}
由{{post.author}}在{post.published}上写的


{{posts.content}
{%endblock%}

post.html的列表

{% extends 'blog/post/base.html' %}

{% block title %}List of blog post{% endblock %}

{% block content %}
     {% for posts in post %}
         <h2><a href="{{posts.get_absolute_url}}">{{ posts.title }}</a></h2>
         <p>Written by {{ posts.author }} on {{ posts.published}}</p>
         <hr>
         {{ posts.content}}
     {% endfor %}
{% endblock %}

{%extends'blog/post/base.html%}
{%block title%}博客文章列表{%endblock%}
{%block content%}
{post%中的post%为%s}
由{{posts.author}}在{{posts.published}上撰写


{{posts.content}
{%endfor%}
{%endblock%}

blog/url.py

from django.conf.urls import url
from django.urls import path
from . import views

app_name = 'blog'
urlpatterns =[
    path('',views.list_of_post,name='list_of_post'),
    path('<slug:slug>/',views.list_of_post,name='post_detail')
]

从django.conf.url导入url
从django.url导入路径
从。导入视图
应用程序名称='博客'
URL模式=[
路径（“”，views.list_of_post，name='list_of_post'），
路径（'/'，视图.列表，名称='post\u detail'）
]

---

如果这太简单了，我很抱歉。但我还是一个程序员初学者。需要一些帮助。谢谢。

在您的

url.py

中，您写道：

from django.conf.urls import url
from django.urls import path
from . import views

app_name = 'blog'
urlpatterns =[
    path('',views.list_of_post,name='list_of_post'),
    path('<slug:slug>/',views.list_of_post,name='post_detail')
]

---

通过将post的列表链接到第二个路径（…）Django将使用

slug

参数调用此函数，但该函数当然无法处理此问题，因此

TypeError

您可能复制了第二个路径，并且无法更改视图（到

post\u detail

）.正确！！这似乎是个打字错误。但我自己都没注意到。非常感谢！！！

from django.conf.urls import url
from django.urls import path
from . import views

app_name = 'blog'
urlpatterns =[
    path('',views.list_of_post,name='list_of_post'),
    path('<slug:slug>/',views.post_detail,name='post_detail')
]