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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