I tried to apply mptt in my django program on a existing model using the method provided on the main site of mptt, which is as follow:
import mptt
from mptt.fields import TreeForeignKey
from django.contrib.auth.models import Group
# add a parent foreign key
TreeForeignKey(Group, blank=True, null=True, db_index=True).contribute_to_class(Group,'self')
mptt.register(Group, order_insertion_by=['name'])
However, when I open my group list in the admin-site, it says that the group model doesn't have such column named parent_id, I wonder how I can fix it.
BTW the code have been written in models.py,is it possible that I should have written it in admin.py?
reference: Registration of existing models
Edit:
sorry about the comment...migrations does solve the problem ><
However it leads to another question...can I only create a group by code to create a new tree structure, or it's possible for me to do that somewhere in my admin site?
Thanks for the answering and attention:)
Related
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Guys I hope all of you are doing great,I have a problem in django3,I have been given a problem and I have to solve it ,I have tried it but not able to solve this,see the image I have given link,
<----question starts here
This list of help1, help2,... gets stored in a database table. You could do a first version of models.py and enable the admin interface for it.
----->question ends here
Can anyone give me a idea?or how to do it or in more simpler words?any ideas or suggestions is appreciated
I'm not really understand what you mean but I think you need to create another model, let's call it Help. After created the Help model you can use it as foreignkey inside your old model.
models.py
# new model for help types
class HelpType(models.Model):
type = models.CharField(max_length=100)
def __str__(self):
return self.type
# old model
class OldModel(models.Model):
# other fields
help_type = models.ForeignKey(HelpType, on_delete=models.CASCADE)
admin.py
# register it to admin panel
from django.contrib import admin
from .models import HelpType
admin.site.register(HelpType)
Now you can create all help types in admin panel and then when you create your OldModel's object all help types will be listed as drop down in help_type field
I want to develop a server side of an app that holds users.
Of course I need a table in database holding the user information.
At first I may write
class User(models.Model): # using django models
userid = ...
password = ...
which gives me a database table containing userid and password.
However, I might want to add some attributes (maybe Credit, Birthday...so on) to each user in the future. I just can't think up all of them right now. And I can't know what attributes I would really need in the future.
How can I deal with it?
There's already a user table in Django. This table is automatically create when you first apply the migration with 'manage.py migrate' command.
In database schema, this table is listed as auth_user and you can import it into Django with the following command
from django.contrib.auth.models import User
Django provides a default model for User. you can use it like this.
from django.contrib.auth.models import User
and as per your second query. you can do so by creating another model and adding a ForeignKey or OneToOneField of User model to link it with each user.
class Customuserprofile(models.Model):
user=models.OneToOneField(settings.AUTH_USER_MODEL)
credit=models.CharField()
birthday=models.DateTimeField()
well, if you want to add some field to the User you can use AbstractUser or AbstractBaseUser model read this that explain the differences and give a example https://simpleisbetterthancomplex.com/tutorial/2016/07/22/how-to-extend-django-user-model.html#abstractbaseuser
(Django 1.10.) I'm trying to follow this advice on extending the user model using OneToOneField. In my app 'polls' (yes, I'm extending the app made in the 'official' tutorial) I want to store two additional pieces of information about each user, namely, a string of characters and a number.
In my models.py I now have the following:
from django.contrib.auth.models import User
class Employee(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
stopien = models.CharField(max_length=100)
pensum = models.IntegerField()
and in admin.py the following:
from django.contrib.auth.admin import UserAdmin as BaseUserAdmin
from django.contrib.auth.models import User
from polls.models import Employee
class EmployeeInline(admin.StackedInline):
model = Employee
can_delete = False
verbose_name_plural = 'employee'
class UserAdmin(BaseUserAdmin):
inlines = (EmployeeInline, )
admin.site.unregister(User)
admin.site.register(User, UserAdmin)
When adding a user using the admin panel my two new fields display correctly. However, when I click 'save', or if I don't add any user and just click on the name of my sole admin user in the admin panel, I get the following error:
OperationalError at /admin/auth/user/1/change/
no such table: polls_employee
I see some questions and answers related to similar problems, but they seem to be relevant for older version of Django. Could anyone give me a tip as to what I should do? Ideally I'd want my two additional fields display in the admin panel, though I suspect this might be a task for the future.
I have to confess I do not understand this paragraph from the documentation just following the advice I'm using:
These profile models are not special in any way - they are just Django models that happen to have a one-to-one link with a User model. As such, they do not get auto created when a user is created, but a django.db.models.signals.post_save could be used to create or update related models as appropriate.
Do I need to tie this 'post-save' to some element of the admin panel?
I'd be very greatful for any help!
You need run makemigrations to create a migration for your new model, and then migrate to run the migration and create the database table.
./manage.py makemigrations
./manage.py migrate
I have a model created in Django 1.5 as below:
class Number(models.Model):
phone_number = models.CharField("Phone Number", max_length=10, unique=True)
I set up Django admin as below:
from django.contrib import admin
from demo.models import Message, Number, Relationship, SmsLog
class NumberAdmin(admin.ModelAdmin):
search_fields = ['phone_number']
admin.site.register(Number, NumberAdmin)
I believe Django add "date_created" column to the database automatically (because I know it sorts the data entries by creation time in admin console). Is there a way to view those time/dates in admin console? The closest I have go to is Django tutorial and StackOverflow ,but I do not want to create another column on my own (pub_date in Django official tutorial's example) and add it if possible. Is there a way to do it and if so, could someone show me how to? Thank you!
Django does not automatically add a date_created column. If you want to track the creation date, you have to declare it in your model.
You may be getting the illusion that it does because if you do not specify a sort order in the model or in the admin class for the model, it will default to sorting by primary key, which will increase according to the order the model instances were created.
edit: I wasn't clear before, I am saving my object in the django admin panel, not in a view. Even when I save the object with no many-to-many relationships I still get the error.
I have a model called TogglDetails that has a ForeignKey relationship with the standard django User model and a MayToManyField relationship with a model named Tag. I have registered my models with django admin but when I try to save a TogglDetails instance I get the error in the title.
Here are my models:
class Tag(models.Model):
name = models.CharField(max_length=30)
def __unicode__(self):
return self.name
class TogglDetails(models.Model):
token = models.CharField(max_length=100)
user = models.ForeignKey(User)
tags = models.ManyToManyField(Tag, blank=True, null=True)
def __unicode__(self):
return self.user.username
class Meta:
verbose_name_plural = "toggl details"
As far as I can tell, there should be no issues with my models and django admin should just save the instance without any issues. Is there something obvious that I have missed?
I am using Django 1.3
The answer to my question was this: Postgres sequences without an 'owned by' attribute do not return an id in Django 1.3
The sequences in my postgres database did not have the "Owned by" attribute set and so did not return an id when a new entry was saved to the db.
As stated by other users:
Postgres sequences without an 'owned by' attribute do not return an id in Django 1.3
The sequences in my postgres database did not have the "Owned by" attribute set and so did not return an id when a new entry was saved to the db
In addition:
This is most likely caused by a backwards incompatible change that renders some primary key types in custom models beyond reach for Django 1.3. See Django trac tickets https://code.djangoproject.com/ticket/13295 and http://code.djangoproject.com/ticket/15682 for more information.
I solved the problem by running the follow commands for the affected tables/sequences.
Specifically running the command:
manage.py dbshell
ALTER SEQUENCE tablename_colname_seq OWNED BY tablename.colname;
change tablename_colname_seq and tablename.colname
Don't let us guess and add the Error message to your question, this gives most information about where it fails.
Have you imported the User model?
from django.contrib.auth.models import User
I've had this problem as well and the only thing I could do was make the M2M fields blank and not set them until I hit Save and Continue Editing.
I think this just may be a framework wart, as you will notice the User section of the Admin site also has a very strict "You can only edit these fields until you save the model".
So my recommendation is to adopt that scheme, and hide the M2M form field until the model has a Primary Key.
I tried Django 1.3 using CPython, with different database setups. I copy-pasted the models from the question, and did some changes: first I added
from django.contrib.auth.models import User
at the top of the file and I put the reference to Tag between quotes. That shouldn't make any difference. Further, I created the following admin.py:
from django.contrib import admin
import models
admin.site.register(models.Tag)
admin.site.register(models.TogglDetails)
For Sqlite3, the problem described doesn't occur, neither for MySQL. So I tried PostgreSQL, with the postgresql_psycopg2 back end. Same thing: I can't reproduce the error.
So as far as I can figure, there's nothing wrong with the code in the question. The problem must be elsewhere.