简单搜索会引发错误django

时间:2017-02-23 21:26:00

标签: python django

我正在尝试我的第一个Django应用程序。以前我使用Flask构建了以下应用程序,但它确实有效。

我希望我的应用搜索github并返回前5个搜索项。

我的应用程序名为assign1。 assign1 / views.py如下

from django.shortcuts import render
from django.template import loader
from .application import getresult

# Create your views here.
from django.http import HttpResponse

def index(request):
    return render(request, 'index.html')

def search(request):
    search_term = request.GET['search_term']
    t = loader.get_template('result.html')
    c = Context({'items':getresult(search_term),'search_term':search_term,})
    return HttpResponse(t.render(c))

assign1 / urls.py如下:

from django.conf.urls import url

from . import views

urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^result/$', views.search, name='search'),
]

我的assign1 / templates / index.html

    <!DOCTYPE html>

    {% block content %}
    <div class = "container">
    <h1>Search Github</h1>
            <form action='/assign1/result'>
                <input type='text' name='search_term'><br>

                <br><br>
                <input type='submit' value='Submit'>
            </form>
    {% endblock %}

我的result.html如下:

<!DOCTYPE html>
<head><title>Github Navigator</title></head>

<body>
    {% block content %}
    <h1>{{search_term}}</h1>
    {% for rep in items %}
        <h2>{{loop.index}} {{rep["respository_name"]}}</h2>
        <h3> Created {{rep["created_at"]}}</h3>
        <a href="{{rep["owner_url"]}}"><img src="{{rep["avatar_url"]}}" alt="avatar" height="42" width="42"/></a>
        {{rep["owner_login"]}}
        <h3>LastCommit</h3>
        {{rep["sha"]}} {{rep["commit_message"]}}  {{rep["commit_author_name"]}}
        <hr/>
    {% endfor %}
    {% endblock %}
</body>

索引打开正常。当我点击按钮时,网站网址为http://127.0.0.1:8000/result?search_term=arrow 我收到一条错误,说找不到页面

Using the URLconf defined in assignment.urls, Django tried these URL patterns, in this order:
^assign1/
^admin/
The current URL, result, didn't match any of these

我的主要urls.py如下:

from django.conf.urls import include, url

from django.contrib import admin

urlpatterns = [
   # Examples:
   # url(r'^$', 'myproject.views.home', name = 'home'),
   # url(r'^blog/', include('blog.urls')),
   url(r'^assign1/', include('assign1.urls')),
   url(r'^admin/', admin.site.urls),
]

我做错了什么?

0 个答案:

没有答案