Django-MPTT子页面的完整路径如何制作? [英] Django-MPTT full path to child pages how to make?

问题描述

我开始使用Django-MPTT应用程序在Django站点页面上获取基于树的方法.

I'm start using Django-MPTT app to get a tree-based approach on my Django-site pages.

例如我的页面带有子页面:

For ex. I have pages with sub pages:

Tr:

• 发声Tra(子页面)
• 硬Tra(子页面)

断裂:

• Atmo Breaks(子页面)
• 渐进式休息(子页面)

如何从urls.py访问它们? 什么模式会有所帮助? 我需要在模型中存储Full_path还是可以通过url模式来完成?

How can I get access to them from urls.py? What pattern will help? Do I need to store Full_path in model or it can be done via url pattern?

推荐答案

我假设您是说要执行以下URL:

I assume you mean you want to do URLs like this:

/trance/
/trance/vocal-trance/
/trance/hard-trace/
/breaks/
/breaks/atmo-breaks/
/breaks/progressive-breaks/

如果是这样，最好将url片段存储在模型中.像这样:

If so, it's probably best to store the url fragment in your model. Something like:

from mptt.models import MPTTModel
from django.db import models
from django.template.defaultfilters import slugify

class Page(MPTTModel):
    name = models.CharField(max_length=50)
    slug = models.CharField(max_length=50,null=True)
    url = models.CharField(max_length=255,null=True)

    def save(self, *args, **kwargs)
        if self.slug is None:
            # create a slug that's unique to siblings
            slug = slugify(self.name)
            self.slug = slug
            siblings = self.get_siblings()
            i = 1
            while siblings.filter(slug=self.slug).exists():
                i += 1
                self.slug = slug + '-%d' % i

            # now create a URL based on parent's url + slug
            if self.parent:
                self.url = '%s/%s' % (self.parent.url, self.slug)
            else:
                self.url = self.slug
        super(Page, self).save(*args, **kwargs)

然后添加一个网址格式:

Then add a URL pattern:

(r'^pages/(?P<page_url>[\w\d_/-]+)/$', 'pages.views.show_page'),

在您看来，您只需获取正确的页面即可

And in your view you can just fetch the right page:

def show_page(request, page_url=None):
    page = get_object_or_404(Page, url=page_url)
    ...

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