如何在Wa中创建不可见的虚拟页面? [英] How to create an invisible dummy page in Wagtail?

问题描述

如何在Wa中创建不可见的虚拟页面?

How can I create an invisible dummy page in Wagtail?

我需要在Wagtail中创建一个虚拟"页面对象，以便为基于非Wagtail的页面以及外部资源构建菜单. (请参阅我的帖子

I need a "virtual" page object within Wagtail to build a menu for non Wagtail based pages and also external resources. (See my entry post here)

class MenuDummyPage(Page):

    menu_description    = models.CharField(max_length=255, blank=True)
    menu_icon           = models.CharField(max_length=255, blank=True)
    menu_link           = models.CharField(max_length=255, blank=True)

    settings_panels = [
        FieldPanel('menu_description'),
        FieldPanel('menu_icon'),
        FieldPanel('menu_link'),
    ]

    def get_sitemap_urls(self):
        return []

    def serve(self, request):
        pass

如果我创建了上述页面对象，则该对象未在生成的wagtail网站地图中列出.

If I create the above page object then it is not listed within the generated wagtail sitemap.

但是，如果我自己手动导航到该页面，则将调用该对象.我该如何阻止呢?

But if I navigate on my own to that page manually the object is called. How can I stop this?

示例: 如果我创建一个标题为"This is a test"的MenuDummyPage，则系统将自动生成一个slug =>"this-is-a-test".

Example: If I create a MenuDummyPage with the title "This is a test" then the system will automatically generate a slug => "this-is-a-test".

如果我在浏览器中调用"/this-is-a-test"/，则ag会回答，因为the存在.如何为我的"MenuDummyPage"对象删除此行为?

If I call "/this-is-a-test"/ in my browser wagtail is answering because the slug exists. How can I remove this behavior for my "MenuDummyPage" objects?

推荐答案

如果虚拟页面的意思是在页面树中保留了其他页面的页面，则可以执行以下操作:

If what you mean by a dummy page is a page under which other pages are kept in the page tree, then you could do the following:

from django.http import HttpResponseRedirect

class Node(Page):

    subpage_types = [your subpage types]
    parent_page_types = [your parent page types]

    link = models.CharField(max_length=255, default='', blank='True')

    content_panels = Page.content_panels + [
        FieldPanel('link')
    ]  

    def serve(self, request):
        if self.link is not None:
            return HttpResponseRedirect(self.link)
        # To handle the situation where someone inadvertantly lands on a parent page, do a redirect
        first_descendant = self.get_descendants().first()
        if first_descendant:
            return HttpResponseRedirect(first_descendant.url)
        else:
            return HttpResponseRedirect(request.site.root_page.url)

可选的link字段允许您在页面树结构中的该位置定义链接.再次，以上假设您将基于Page的项目用作Page树中的占位符，以便可以在其下放置其他Page.只要您没有在模板中呈现此页面的URL，那么用户就永远不会知道如何获取Node的URL，但是如果有人确实获取了Node类型的页面的URL，则，然后first_descendant逻辑通过将它们发送到Node的第一个后代或如果没有Node的后代发送到首页来处理这种情况.

The optional link field allows you to define a link if you wish for this position in the page tree structure. The above, again, assumes that you are using this Page-based item as a placeholder in the Page tree so that you can have other Pages under it. As long as you don't render the url of this page in your template, then users should never know how to get to the url for the Node, but if someone does get to the url for a Node-type page, then the first_descendant logic handles this situation by sending them to either the first descendant of the Node or to the home page if no descendants of Node exist.

在模板中(请注意使用 specific_class ):

In your template (note the use of specific_class):

{% for item in menu_items %}
    <li>
        <a href="{% if item.specific.link and item.specific.link != '' %}{{ item.specific.link }}{% elif item.specific_class == 'Node'%}#{% else %}{% pageurl item %}{% endif %}">{{ item.title }
        </a>
    </li>
{% endfor %}

这篇关于如何在Wa中创建不可见的虚拟页面?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！