Hello, Shai and Edgar

yes, I've tried using *inspectdb*, however, it only generates models for 
database tables, but not for database View[1] and database Synonym[2].

The Django documentation is currently not explicit in showing that the 
specific database features are supported by Django database backends.

*inspectdb *does not support "database schemas"[3][4].

Any other ideas how to solve this?


[1] 
http://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_8004.htm
[2] 
http://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_8004.htm
[3] https://groups.google.com/d/msg/django-users/jsoDvI7DipU/CYOdstKeTYcJ 
[4] https://code.djangoproject.com/ticket/22673


-- 
Fábio C. Barrionuevo da Luz
Acadêmico de Sistemas de Informação na Faculdade Católica do Tocantins - 
FACTO
Palmas - Tocantins - Brasil - América do Sul

http://pythonclub.com.br/


Em quinta-feira, 1 de janeiro de 2015 16h57min38s UTC-3, Edgar Gabaldi 
escreveu:
>
> Fabio,
>
> Django ORM support Oracle Database[1].
>
> If you want connect an existing oracle database, i recomend you see the 
> inspectdb management command.
>
> [1] https://docs.djangoproject.com/en/1.7/ref/databases/#oracle-notes
> [2] https://docs.djangoproject.com/en/1.7/howto/legacy-databases/
>
>
> On Thu, Jan 1, 2015 at 2:40 PM, Shai <sh...@platonix.com <javascript:>> 
> wrote:
>
>> Hi Fábio,
>>
>> On Wednesday, December 31, 2014 6:05:05 PM UTC+2, Fabio Caritas 
>> Barrionuevo da Luz wrote:
>>>
>>> Hello, is possible with Django 1.7 create a unmanaged Django model for 
>>> Oracle database View?
>>>
>>>
>>>
>> Yes. In your model's Meta, set managed to False and db_table to the view 
>> name:
>>
>> class MyUnManagedModel(models.Model):
>>     # ...
>>     # fields
>>     # ...
>>     class Meta:
>>         managed = False
>>         db_table = 'view_name'
>>
>> Have you run into difficulties?
>>
>> HTH,
>> Shai.
>>
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