Customizing django object parameters value output - django

Question

what does django do, when i do something like that in template
{{ object.parameter }}
I ask this because in case of arrayfields (postgresql array fields) it will print out either
{'value', 'value', 'value'}
(because thats how postgresql stores arrays in arrayfields)
or
['value','value','value']
if i use fields post_init method to convert postgresql array to python list.
Desired output would of course be value, value, value. I would rather not use some kind of filters for that, because then i would have to resort using IFs in templates or use some kind of template tag filter for every value i print out and that does not feel like a smart thing to do.
By the way, i know i can do something like that in template :
{% for choice in field.choices %}
{{ choice }}
{% if forloop.last %}
{% else %},
{% endif%}
{% endfor %}
and that gives me exactly what i want, but i thought there would be some other way doing it - with some modelfield method or something.
Alan

Answer 1

what does django do, when i do something like that in template
{{ object.parameter }}
See variables and lookups.
Desired output would of course be value, value, value. I would rather not use some kind of filters for that, because then i would have to resort using IFs in templates or use some kind of template tag filter for every value i print out and that does not feel like a smart thing to do.
You can make a really trivial filter:
#register.filter
def comma_join(values):
return u', '.join(values)
So simple:
{{ object.parameter|comma_join }}
Why would you want to avoid such a simple solution ?
and that gives me exactly what i want, but i thought there would be some other way doing it - with some modelfield method or something.
Of course you could also add such a method:
class YourModel(models.Model):
# ....
def comma_join_parameter(self):
return u', '.join(self.parameter)
And use it in your template as such:
{{ object.comma_join_parameter }}