登录Django后根据用户名将用户重定向到自定义url

Question

Hello? I am trying to redirect a particular user to a custom page once they log in in django. The Admin will be directed to their usual admin interface while this particular user will go to their own custom page. I have written this code and placed it in my views

def custLogin(request):
    if request.user.username == '***':
        return HttpResponseRedirect('http://********************.html')
    else:
        return login(request,template_name='login.html')

I have pointed the accounts/login url in urls.py to custLogin as below

(r'^accounts/login/', custLogin),

I however keep getting the error

Caught NoReverseMatch while rendering: Reverse for 'django.contrib.auth.views.login' with arguments '()' and keyword arguments '{}' not found.

Any pointers please?

Answer 1

Maybe, you should use redirect shortcut, which returns an HttpResponseRedirect , to point your users to different places after custLogin view processed.

from django.shortcuts import redirect
def custLogin(request):
    if request.user.is_staff:
        return redirect('/some/admin/url')
    else:
        return redirect('/some/regular/user/url')

As documentation says, you can use the redirect() function in a number of ways: by passing a view name, an object or url. See documentation for more info

Note, that I use is_staff user field to determine: is user an admin or not.

Answer 2

In your template, you may have forgotten to do this at the beginning of your template:

{% load from future url %}

This is useful when using:

{% url 'name_of_my_url_inurl.py' %}

Answer 3

For redirecting each user to his own page use reverse

First create url route which accepts parameter. It's described here

And redirect like

return HttpResponseRedirect(reverse('arch-summary', args=[1945]))

where arch-summary - named url, and in list args - parameters for it. Examplehere

To distinguish admin user from usual user, check it in view and redirect to different url.