Sending argument in reverse_lazy django - Python -

i using reverse_lazy following

def index:     if firsttimelogin == 0:          response = httpresponseredirect(reverse_lazy('abc', args={'cid': 0}))     else:          response = httpresponseredirect(reverse_lazy('abc', args={'cid': 1})) 

abc view

def abc(request, userid = none, cid =1 ):     return httpresponse(cid) 

urls.py

url(r'^abc/$', views.abc, name='abc'), url(r'^abc/(?p<userid>.*)/$', views.abc, name='abc'), url(r'^abc/(?p<cid>.*)/$', views.abc, name='abc'), 

now when redirected def abc through reverse_lazy url like: baseurl/appname/abc/cid, here dont know how can fetch value of cid.

please correct if other approach required. need pass argument reverse_lazy want fetch in def abc

update

refering romaan saying: have updated: urls.py as:

url(r'^abc/(?p<cid>.*)/$', views.abc, name='abcfirst') 

and redirect as:

response = httpresponseredirect(reverse_lazy('abcfirst', args={'cid': 1})) 

still unable access arguments sending through reverse_lazy.

you have 2 urls have same name, , take in same parameters (same regex).

you can't combine these 2 have view take in both, thats not how django works.

you either have to

• make url takes in 2 parameters
• make separate views

but either way, can't use same name different urls.