I'm not sure, but I think this should work - try it out and let us know. r'^user/([\w\-+.@]+)/$'
I'm not too good with regexes, either. No one is. Also, characters like "+" are encoded in URLs - so you might want to think of that. On Sun, Apr 22, 2012 at 1:42 PM, Swaroop Shankar V <swaroo...@gmail.com>wrote: > Hello All, > I need some urgent help with this one. Am trying to show a user profile > when the user navigates to a url of this format > > http://example.com/user/<username> > > I have provided the following expression in my urls.py file > > (r'^user/(\w+)/$','accounts.views.userProfile'), > > this working fine for normal usernames, but if the user add a . (dot) in > between his username the url do not match with the regex and throws Page > not found error. > > Since the django username supports Letters, digits and @/./+/-/_ i want > all these to be supported in my url. Could someone please help me to > construct the url. Am pretty weak in regex and it will take me sometime to > build a proper regex for the above url. Thanks > > Regards, > > Swaroop Shankar V > > -- > You received this message because you are subscribed to the Google Groups > "Django users" group. > To post to this group, send email to django-users@googlegroups.com. > To unsubscribe from this group, send email to > django-users+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/django-users?hl=en. > -- Yati Sagade <http://twitter.com/yati_itay> Twitter: @yati_itay <http://twitter.com/yati_itay> Organizing member of TEDx EasternMetropolitanBypass http://www.ted.com/tedx/events/4933 https://www.facebook.com/pages/TEDx-EasternMetropolitanBypass/337763226244869 -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
