Django making a custom 403 page

def csrf_failure(request, reason=""):    
  context = RequestContext(request)    
  response = render_to_response('error403.html', context)    
  response.status_code = 403    
  return response

Source: evileg.com

Similar codes

0 vote142 views1 answer
Loop through words in a string python
Python
0 vote172 views1 answer
Pygame mixer documentation
Python
0 vote172 views2 answers
Copy only some columns to new dataframe in r
Python
0 vote198 views1 answer
Delete history django simple
Python Django
0 vote177 views1 answer
Python 123 + 0.00
Python
24 votes172 views5 answers
Remove a part of a string python
Python
2 votes181 views1 answer
Extract float from string python
Python
0 vote184 views1 answer
Python kwargs get
Python
1 vote191 views3 answers
Recursive function
Python
0 vote171 views1 answer
If between two values pythpn
Python