Question

我在Django中的index.html模板具有以下内容：

<ul>
    <li><a href="{% url 'quantities' %} value="1">All</a></li>
    <li><a href="{% url 'quantities' %} value="2">Subset1</a></li>
    <li><a href="{% url 'quantities' %} value="3">Subset2</a></li>
    <li><a href="{% url 'quantities' %} value="4">Subset3</a></li>
</ul>

因此链接到相同的模板，但（取决于点击的链接）我想写我的app / views.py中的函数：

def quantities(request):
    if value == "1":
        levels = Model_1.objects.all()
    elif value == "2":
        levels = Model_1.objects.filter(createria=2)
    elif value == "3":
        levels = Model_1.objects.filter(createria=3)
    elif value == "4":
        levels = Model_1.objects.filter(createria=4)

我可以获取上一页中点击链接的值吗？

Answer 1

我最好传递url中的值而不是url参数：

<ul>
    <li><a href="{% url 'quantities_all' %}>All</a></li>
    <li><a href="{% url 'quantities' '2' %}>Subset1</a></li>
    <li><a href="{% url 'quantities' '3' %}>Subset2</a></li>
    <li><a href="{% url 'quantities' '4' %}>Subset3</a></li>
</ul>

您的观点将如下所示：

def quantities(request, value=None):
    if value in ('2', '3', '4'):
        levels = Model_1.objects.filter(createria=int(value))
    else:
        levels = Model_1.objects.all()
    ...

不要忘记改变你的网址：

url(r'^quantities/$', 'app.views.quantities', name='quantities_all'),
url(r'^quantities/(\d+)/$', 'app.views.quantities', name='quantities'),

Answer 2

将值作为GET参数传递：

<ul>
    <li><a href="{% url 'quantities' %}?value=1">All</a></li>
    <li><a href="{% url 'quantities' %}?value=2">Subset1</a></li>
    <li><a href="{% url 'quantities' %}?value=3">Subset2</a></li>
    <li><a href="{% url 'quantities' %}?value=4">Subset3</a></li>
</ul>

获取视野：

def quantities(request):
    value = request.GET.get('value', '1')
    if value == "1":
        levels = Model_1.objects.all()
    elif value == "2":
        levels = Model_1.objects.filter(createria=2)
    elif value == "3":
        levels = Model_1.objects.filter(createria=3)
    elif value == "4":
        levels = Model_1.objects.filter(createria=4)

Answer 3

您需要在网址配置中捕获数量。 https://docs.djangoproject.com/en/1.7/topics/http/urls/

这比传递GET参数更好，因为你有一个链接，像http://example.org/quantities/4/这样的链接看起来更好，而在Django这是一个惯例：

<ul>
    <li><a href="{% url 'quantities' 1 %}">All</a></li>
    <li><a href="{% url 'quantities' 2 %}">Subset1</a></li>
    <li><a href="{% url 'quantities' 3 %}">Subset2</a></li>
    <li><a href="{% url 'quantities' 4 %}">Subset3</a></li>
</ul>

所以你的urls.py看起来像：

from django.conf.urls import patterns, url
from . import views

urlpatterns = patterns('',
    url(r'^quantities/(\d+)/$', views.quantities), )

views.py来自另一个答案

def quantities(request, value):
    if value in ('2', '3', '4'):
        levels = Model_1.objects.filter(createria=int(value))
    else:
        levels = Model_1.objects.all()
    ...

Django视图和模板链接

3 个答案: