ogle.com/d/msgid/django-users/afc98569-1c64-4322-a533-d62a1a297115n%40googlegroups.com.
Environment:
Request Method: GET
Request URL: http://127.0.0.1:8000/user/
Django Version: 4.0.4
Python Version: 3.10.4
Installed Applications:
['django.contrib.admin',
The best you can do is look for a Referrer header in the request:
request.META['HTTP_REFERER']
which should be the URL of the originating page. However that may or may not
be there depending on user's browser etc etc.
Malcolm
On 3 August 2011 00:41, He Jibo wrote:
> Hi, Djangoers,
>
> I am tr
Hi, Djangoers,
I am trying to write a page rank banner. *I need your help on how to get the
request UR*L. I want the function looks like this. If I put the following
html snippet in a URL, for example,
http://www.homesecurity361.com/index.html, I want the following code to show
the page rank of t
gt;
>> On 9 jan, 17:06, "burcu hamamcıoğlu" wrote:
>> > Hi everybody,
>> >
>> > do you know how can i get request url in views.py.
>> >
>> > My url pattern is like
>> (r'^wp/rmain2\.aspx','DisplayPacketApplications2&
Thanks Bruno request.get_full_path() is enough for me .
Best regards
09 Ocak 2009 Cuma 18:31 tarihinde bruno desthuilliers <
bruno.desthuilli...@gmail.com> yazdı:
>
>
>
> On 9 jan, 17:06, "burcu hamamcıoğlu" wrote:
> > Hi everybody,
> >
> > do
On 9 jan, 17:06, "burcu hamamcıoğlu" wrote:
> Hi everybody,
>
> do you know how can i get request url in views.py.
>
> My url pattern is like (r'^wp/rmain2\.aspx','DisplayPacketApplications2'),
> But my link is like :http://wap2.cepoyun.com
Hi everybody,
do you know how can i get request url in views.py.
My url pattern is like (r'^wp/rmain2\.aspx','DisplayPacketApplications2'),
But my link is like :
http://wap2.cepoyun.com/wp/rmain2.aspx?pid=194&w=280jhjh
when i use 'PATH_INFO' i olnly get '
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