Django反向url与参数到基于类的视图

[英]Django reverse url with parameters to a class based view

本文翻译自 user769498 查看原文 2011/12/21 3162 hyperlink/ python/ django/ django-class-based-views

I just started learning python and django and I have a question. I got the assignment to turn function views into class based views. But my links wont work now.

我刚开始学习python和django，我有一个问题。我得到了将函数视图转换为基于类的视图的赋值。但我的链接现在不起作用。

these are from urls.py:

这些来自urls.py：

url(r'^$', ContactIndex.as_view()),
url(r'^add$', ContactAdd.as_view()),
url(r'^([0-9]+)/update$', ContactUpdate.as_view()),
url(r'^([0-9]+)/view$', ContactView.as_view()),

This is my link :

这是我的链接：

{% url rtr_contact.views.ContactView contact.id %}

but this doesnt work it says:

但这不起作用它说：

Caught NoReverseMatch while rendering: Reverse for 'rtr_contact.views.ContactView' with arguments '(20L,)' and keyword arguments '{}' not found.

1 个解决方案

#1

To make url reversing easy, I recommend that you always name your url patterns.

为了使网址翻转变得容易，我建议您始终为网址模式命名。

url(r'^$', ContactIndex.as_view(), name="contact_index"),
url(r'^add$', ContactAdd.as_view(), name="contact_add"),
url(r'^([0-9]+)/update$', ContactUpdate.as_view(), name="contact_update"),
url(r'^([0-9]+)/view$', ContactView.as_view(), name="contact_view"),

Then in the template:

然后在模板中：

{% url contact_view contact.id %}

注意！

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