Thank you;
my main problem was with the sites model - and the default value for
SITE_ID == 1. I had no intention of hardcoding the URL.
On Mon, Jun 11, 2018 at 12:58 AM Anthony Anonde <demeyo...@gmail.com> wrote:
> the example.com is am example django give for Site. Which means there is
> a way for you to manage more than one site at a time, basically using the
> site_id.
> To you Questions, i don't understand why you want to hard code your uri
> routing like that when you can easily use the app_name and name convention
> which is more dynamic then hard coding it like that:
> for example at the model Transaction urls.py
> app_name = "trans"
> url(r'(?P<pk>-/d+)$', views.func, name='account_trans'),
>
> def get_absolute_url(self):
> return reverse("trans:account_trans", kwargs={"pk":self.pk})
>
>
> and you will be just fine.... hope this help
>
>
>
> On Sunday, June 10, 2018 at 12:30:51 PM UTC+1, Joakim Hove wrote:
>>
>> Hello,
>>
>> I have a model `Transaction` which can be viewed at the url:
>> /transaction/view/$ID/ - if I just enter that url in the browser, or in
>> template everything works. But when I have the following get_abolute_url()
>> method:
>>
>> def get_absolute_url(self):
>> return "/transaction/view/{}/".format( self.id )
>>
>> things do not work from the Admin. When I click on the "View on Site"
>> link i am redirected to "http://example.com/transaction/view/23635/"; and
>> I get an error message about failed XML parsing. I have no clue where the "
>> example.com" address comes from - that string is not in my codebase. If
>> I hardcode the get_absolute_url to:
>>
>> def get_absolute_url(self):
>> return "http://127.0.0.1:8000/transaction/view{}/".format(self.id)
>>
>> That works in development, but needless to say that is not a very good
>> approach.
>>
>> I am on Django 1.10
>>
>>
>> Regard Joakim
>>
>>
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