如何使用Django HttpResponseRedirect提供规范的URL？ [英] How to provide canonical URL with Django HttpResponseRedirect?
问题描述
这个问题与我刚问过的一个问题非常相似 href：我可以获取/显示最终重定向URL的Google搜索结果吗？，但现在的问题只针对Django。
This question is very similar to one I just asked href: Can I get Google search results to use/display the final redirect url?, but now the question is specific to Django.
我的网站的网页网址使用以下格式：
My site has webpage urls that use the following format:
www.mysite.com/id/pretty_title
首页链接到这些页面，但href实际上包含一些参数：
The front page links to these pages, but the href actually contains some parameters:
www.mysite.com/id/?some_ugly_parameters_to_let_me_know_what_search_it_is_from
然后重定向到
www.mysite.com/id/pretty_title
显示页面。
我的问题是 Google的搜索结果显示页面链接为丑陋的URL，而不是经过重定向的URL。
My issue is that Google's search results show the link to the page as the ugly url instead of the pretty redirected one.
我了解到的是，我需要提供一个规范的链接。但是，当丑陋的url页面从未真正存在时，至少不像我所写的那样，该怎么办？
What I have learned is that I need to provide a canonical link. But how can I do this when the ugly url page never really exists, at least not as one that I have written?
服务器端发生的事情是丑陋的网址会重定向：
What happens server side is that the view of the ugly url does a redirect:
return HttpResponseRedirect(pretty_url)
推荐答案
您可以将其作为Django模板返回的HTML的一部分，放在<头> 部分。
您的Django中是否有 base.html ？您可以将 {％block％} 设置为规范URL的占位符，然后在 {％扩展基数的每个页面中设置该值。 html％}
You can just put it as part of the HTML returned from the Django template, in the <head> section.
Do you have a base.html in your Django? You can setup a {% block %} as a placeholder for the canonical URL and then set that value in each individual page that {% extends base.html %}
base.html
base.html
<html>
<head>
<link rel="canonical" href="{% block canonical_url %}{% endblock %}">
</head>
...
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