SyntaxError: URLpattern 中的语法无效

Question

hi am getting a syntax error

url:

url(r'^reset-password/$',
    PasswordResetView.as_view(template_name='accounts/reset_password.html', 'post_reset_redirect': 'accounts:password_reset_done'), name='reset_password'),

What is the problem?

thanks

Answer 1

The problem is that you mix dictionary syntax with parameter syntax:

url(
    r'^reset-password/$',
    PasswordResetView.as_view(
        template_name='accounts/reset_password.html',
        
    ),
    name='reset_password'
)

This syntax with a colon, is used for dictionaries. For parameters, it is identifier=expression , so:

from django.urls import 

url(
    r'^reset-password/$',
    PasswordResetView.as_view(
        template_name='accounts/reset_password.html',
        
    ),
    name='reset_password'
)

The post_reset_redirect has been removed as parameter, but the success_url performs the same functionality: it is the URL to which a redirect is done, after the POST request has been handled successfully.

The wrong syntax probably originates from the fact that when you used a function-based view , you passed parameters through the kwargs paramter, which accepted a dictionary.

The class-based view however, obtains these parameter through the .as_view(..) call. Furthermore class-based views typically aim to generalize the process, and there the success_url , is used for FormView s.