My product model has two different relationship with category model:
1. One to One
2. One to Many through ProductShadowCategory table.
Now the situation is when I tried to fetch using the second relationship, I
am getting the result from my first relationship.
For example this is what I
to have a select field with 2000 records I want it to be only reading to
show me the field to which is linked only reading
if I put it in disabled mode I deactivate the ok field to this point well,
the bad thing is that putting it this way, the 2000 records are still
loaded into the template
a possible solution I see is to change the field to CharField() type, here
is a new problem: the field is printed the pk that is related as then how
can I read that pk to show the name and not that pk in the template?
El martes, 18 de septiembre de 2018, 13:58:26 (UTC-5), nelson fernando
to have a select field with 2000 records I want it to be only reading to
show me the field to which is linked only reading
if I put it in disabled mode I deactivate the ok field to this point well,
the bad thing is that putting it this way, the 2000 records are still
loaded into the template
The latest django version you can use for mongodb is 1.6
Flask another framework you use instead of Django, and it support mongodb.
On Mon, Sep 17, 2018 at 4:16 AM akshat ahuja
wrote:
> Hey guys,
> I am a beginner to Django and I am making an application in which I have
> to connect to MongoDB
Your problem is caused by not properly structurizing your url files. An
example (just for reference) is shown below:
*/myappointments/urls.py*
# includes...
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('clinic.urls'), namespace='clinic'), # for
example.com/ where
Sorry but I don't understand. For me, what you say now is that you want the
field to be visible but empty (without any data from model in it) or
invisible (excluded)? Can you, please, clarify what do you want to achieve?
Have you tried appending the lines I gave you, not replacing the last line?
I think this answers my question , trying to implement it.
On Wednesday, September 19, 2018 at 4:45:02 AM UTC+5:30, Mateusz wrote:
>
> In that answer I assume Jobs are connected anyhow with Persons with a
> ForeignKey field like so:
>
> *File /appname/models.py:*
>
> from django.db import models
Reference : https://www.youtube.com/watch?v=6lafzyjiqgQ
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Hey Mike,
Please provide the full traceback.
Also, since all your conditions are ANDed you could use a dict to build
filter() kwargs instead of Q() objects.
Cheers,
Simon
Le mercredi 19 septembre 2018 10:36:13 UTC-4, Matthew Pava a écrit :
>
> I’m not entirely sure if the error is from your Q
I’m not entirely sure if the error is from your Q chain, but something I would
try is to replace “[:1].get()” with “.first()”.
wm = Boat.objects.filter(name=x).filter( f ).first()
From: django-users@googlegroups.com [mailto:django-users@googlegroups.com] On
Behalf Of MikeKJ
Sent: Wednesday,
Using this as an example from Dan Herrar
enter code here
from django.db.models import Q
user_pk = 1
category_pk = 1 #some times None
f = Q( user__pk = user_pk, date=now() )
if category_pk is not None:
f &= Q( category__pk = category_pk )
This only disables it but internally it receives data this makes the
process of loading the template is the same with disabled enabled or
disabled, so it is necessary to use readonly but does not show readonly.
El martes, 18 de septiembre de 2018, 19:00:37 (UTC-5), Mateusz escribió:
>
> (use
This only disables it but internally it receives data this makes the
process of loading the template is the same with disabled enabled or
disabled, so it is necessary to use readonly but does not show readonly.
El martes, 18 de septiembre de 2018, 19:00:37 (UTC-5), Mateusz escribió:
>
> (use
This is not possible with the mysql database, as mysql doesn't support
schemas. Schemas in mysql creates multiple databases - this means that you
will have to create a new database for each client. Postgres is a much
better solution for these types of things.
Regards,
Andréas
Den ons 19 sep.
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On Wed, Sep 19, 2018 at 12:09:20AM -0700, Devender Kumar wrote:
> Hi,
> I need a help from you I wanna start a project which is a multitenant
> application and I want the tenant can register themselves and schema is
> created for him from the
HI Everett White , Please suggest solution i am trying since 3 days, I want
Django model update if exist. I will very thankful to you.
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Hey y’all
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Follow the django tutorial. Once you complete part 2, you can easily start
writing.
https://docs.djangoproject.com/en/2.0/intro/tutorial01/
You don't have to worry too much about schema as you use models.
On Wed 19 Sep, 2018, 12:39 PM Devender Kumar, wrote:
> Hi,
> I need a help from you I
Hi Akshat,
Could you be more specific? Do you have some code to provide to us in order
to help you? Do you have errors?
Pymongo is a good lib if you intend not to use the Django ORM, if you want
to use the ORM, Djongo exists, or others ODMs
Kind regards,
Benjamin
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All this should be done using the MySQL database.
On Wednesday, September 19, 2018 at 12:39:20 PM UTC+5:30, Devender Kumar
wrote:
>
> Hi,
> I need a help from you I wanna start a project which is a multitenant
> application and I want the tenant can register themselves and schema is
> created
Hi,
I need a help from you I wanna start a project which is a multitenant
application and I want the tenant can register themselves and schema is
created for him from the shared database across tenants.
I want to make this application in Django I know all the terminology of how
multitenancy can
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