Question

如何将绝对URL传递给django函数并重定向到它？

def function(request):
    back_url = request.META['HTTP_REFERER']    # example.com/home/?status=80&page=1
    return redirect(back_url)

Answer 1

你可以这样做：

def func(request):
    url  = request.META['HTTP_REFERER']
    if request.META['QUERY_STRING']:
        url += '?%s' % request.META['QUERY_STRING']
    return redirect_to(request, url, **kwargs)

如何在django中传递绝对URL

1 个答案: