如何在Django中重定向到其他URL? [英] How can I redirect to a different URL in Django?

问题描述

我有两页，一页显示特定项目的详细信息，另一页搜索项目.假设urls.py已为它们都正确配置，并且在我的应用程序的views.py中，我具有:

I have two pages, one which is to display details for a specific item and another to search for items. Let's say that the urls.py is properly configured for both of them and within views.py for my app, I have:

def item(request, id):
    return render(request, 'item.html', data)

def search(request):
    #use GET to get query parameters
    if len(query)==1:
        #here, I want to redirect my request to item, passing in the id
    return render(request, 'search.html', data)

我该怎么做才能正确地重定向请求?我尝试了返回项目(请求，id)，它可以呈现正确的页面，但是它不会更改地址栏中的URL.我如何才能使其真正重定向，而不仅仅是呈现其他页面?项目页面的URL模式为/item/{id} .我已经看过SO和文档中类似问题的答案，但仍无法弄清楚.

What do I do to properly redirect the request? I've tried return item(request, id) and that renders the correct page, but it doesn't change the URL in the address bar. How can I make it actually redirect instead of just rendering my other page? The URL pattern for the item page is /item/{id}. I've looked at the answer to similar questions on SO and the documentation, but still couldn't figure it out.

我不到一个星期前才开始学习Python，另一个答案还不够明确，无法帮助我.

I just started learning Python less than a week ago and the other answer isn't clear enough to help me out.

Nvm，不确定我以前做错了什么，但是当我再次尝试时它起作用了.

Edit 2: Nvm, not sure what I did wrong before, but it worked when I tried it again.

推荐答案

您可以使用

You can use HttpResponseRedirect:

from django.http import HttpResponseRedirect

# ...

return HttpResponseRedirect('/url/url1/')

其中"url"和"url1"等于重定向的路径.

Where "url" and "url1" equals the path to the redirect.

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