如何在django中生成url ?

In Django's template language, you can use {% url [viewname] [args] %} to generate a URL to a specific view with parameters. How can you programatically do the same in Python code?

在Django的模板语言中，可以使用{% url [viewname] [args] %}来生成具有参数的特定视图的url。如何在Python代码中以编程方式执行相同的操作?

What I need is to create a list of menu items where each item has name, URL, and an active flag (whether it's the current page or not). This is because it will be a lot cleaner to do this in Python than the template language.

我需要创建一个菜单项列表，其中每个项都有名称、URL和活动标记(无论它是否是当前页面)。这是因为在Python中这样做要比模板语言干净得多。

3 个解决方案

#1

If you need to use something similar to the {% url %} template tag in your code, Django provides the django.core.urlresolvers.reverse(). The reverse function has the following signature:

如果您需要在代码中使用类似于{% url %}模板标记的东西，Django提供了Django .core.urlresolvers.reverse()。反向函数具有以下签名:

reverse(viewname, urlconf=None, args=None, kwargs=None)

https://docs.djangoproject.com/en/dev/ref/urlresolvers/

#2

I'm using two different approaches in my models.py. The first is the permalink decorator:

我在我的模型中使用了两种不同的方法。第一个是permalink decorator:

from django.db.models import permalink

def get_absolute_url(self): 
    """Construct the absolute URL for this Item."""
    return ('project.app.views.view_name', [str(self.id)])
get_absolute_url = permalink(get_absolute_url)

You can also call reverse directly:

也可以直接调用reverse:

from django.core.urlresolvers import reverse

def get_absolute_url(self): 
    """Construct the absolute URL for this Item."""
    return reverse('project.app.views.view_name', None, [str(self.id)])

#3

Be aware that using reverse() requires that your urlconf module is 100% error free and can be processed - iow no ViewDoesNotExist errors or so, or you get the dreaded NoReverseMatch exception (errors in templates usually fail silently resulting in None).

请注意，使用reverse()要求您的urlconf模块是100%无错误的，并且可以进行处理——iow没有viewnotexist错误，或者您得到了可怕的NoReverseMatch异常(模板中的错误通常是无声地失败，导致无错误)。

#1