I have a django movie model
class Film(models.Model):
title = models.CharField(max_length=200)
movie_id = models.CharField(max_length=8, unique=True, primary_key=True)
director = models.ForeignKey('Director', on_delete=models.SET_NULL, null=True)
year = models.IntegerField(null=True)
genres = models.ManyToManyField(Genre)
I need to use movie_id as primary key, but also i need a field, which represents number of item's row in table.
It must increment automatically, just like standard "id" field.
How can i add it?
This question https://stackoverflow.com/users/3404040/take-care is similar, but i can't use my "number" field as primary key, because movie_id is already used for that.
You can use something like this, but it can be resource consuming if you do not want to use the default id field.
class Film(models.Model):
def number():
no = Film.objects.count()
return no + 1
movie_row = models.IntegerField(unique=True,default=number)
From Django's Documentation:
By default, Django gives each model the following field:
id = models.AutoField(primary_key=True)
This is an auto-incrementing primary key.
If you’d like to specify a custom primary key, just specify primary_key=True on one of your fields. If Django sees you’ve explicitly set Field.primary_key, it won’t add the automatic id column.
Each model requires exactly one field to have primary_key=True (either explicitly declared or automatically added).
If you want yours to explicitly be called movie_id you probably need something like:
class Film(models.Model):
movie_id = models.AutoField(primary_key=True)
Related
class Plans(models.Model):
id = models.IntegerField(primary_key=True)
name = models.CharField(max_length=255)
plan_type = models.CharField(max_length=255)
class Order(models.Model):
id = models.IntegerField(primary_key=True)
selected_plan_id = models.IntegerField(primary_key=True)
Order's selected_plan_id is Plans's id.
Which model should I add a foreign key to? How?
First of all there are some bad ways to pointout:
two fields cannot be primary keys in a table
also django as default includes primary key id in every table, so no need to add id field.
You should be doing this way:
class Order(models.Model):
selected_plan_id = models.ForeignKey(Plans, on_delete=models.CASCADE)
The solution that you are looking for
class Order(models.Model):
id = models.IntegerField(primary_key=True)
selected_plan_id = models.ForeignKey(Plans, on_delete=models.CASCADE)
The purpose of using models.CASCADE is that when the referenced object is deleted, also delete the objects that have references to it.
Also i dont suggest to you add 'id' keyword to your property, django makes automatically it. If you add the 'id' keyword to end of the your property like this case, you gonna see the column called 'selected_plan_id_id' in your table.
class Order(models.Model):
id = models.IntegerField(primary_key=True)
selected_plan_id = models.IntegerField(primary_key=True)
Plain= models.ForeignKey(Plain)
Check the dependence of the table and after getting that made one key as foreign like in this one plain is not depend on the order. But the order depends on the plan.
I am using the command python manage.py makemigrations
However, I get this error:
You are trying to add a non-nullable field 'id' to contact_info
without a default; we can't do that (the database needs something to
populate existing rows). Please select a fix: 1) Provide a one-off
default now (will be set on all existing rows) 2) Quit, and let me
add a default in models.py
Here is models.py:
class Posts(models.Model):
id = models.AutoField(primary_key=True)
post_uesr = models.ForeignKey(User, on_delete=models.CASCADE, null=True, blank=True,related_name='create_user')
post_title = models.CharField(max_length=200)
post_description = models.TextField(max_length=800, null=True, blank=True)
post_likes = models.IntegerField(default=0)
post_date = models.DateTimeField(default=datetime.now)
def __str__(self):
return f'{self.post_uesr}'
It's not a bug, it's documented and logical. You add a new field, which is (by best practice, as you noticed) not NULLable so django has to put something into it for the existing records .
You can press 1 so this option applies
1) Provide a one-off default now (will be set on all existing rows)
so just press 1, if thats your case as value.
else if you want to abort this task and provide manually press 2 and this option applies.
2) Quit, and let me add a default in models.py
BUT here in your case id is a default field already included in every table you create and that will be default to primary key.
And if you are looking to add another AutoField probably this will help
order = models.AutoField(primary_key=False)
And if you are trying to make primary key as uuid4 format
import uuid
id = models.AutoField(default=uuid.uuid4, primary_key=True)
it's telling you should specified a default value for your id field.
ask you for adding manually a default value to this field or make a field without any default value in database.
for example for using UUID as your post id you should do like this:
import uuid
from django.db import models
class Posts(models.Model):
id = models.AutoField(default=uuid.uuid4, primary_key=True)
# other lines of codes
and this is because you are using an AutoField.
you can change it like this:
import uuid
from django.db import models
class Posts(models.Model):
id = models.UUIDField(default=uuid.uuid4, primary_key=True, auto_created=True)
# other lines of codes
I am developing a web app using Django. I have created the table in MySQL database and then generated the models.py using inspectdb. I am able to fetch details and connect to the database without any issues. But while saving the values to the particular table, below error is shown
sav_list = List(id=4, item_name ='name1', item_desc='desc1', location='location', reason='rfp', pid=3)
Cannot assign "3": "List.id" must be a "Order" instance.
my models
class List(models.Model):
id = models.IntegerField(db_column='ID', primary_key=True) # Field name made lowercase.
item_name = models.CharField(db_column='Item_Name', max_length=255) # Field name made lowercase.
item_desc = models.CharField(db_column='Item_Desc', max_length=300) # Field name made lowercase.
location = models.CharField(db_column='Location', max_length=100, blank=True, null=True) # Field name made lowercase.
reason = models.CharField(db_column='Reason', max_length=100, blank=True, null=True) # Field name made lowercase.
pid = models.ForeignKey('Order', models.DO_NOTHING, db_column='PID') # Field name made lowercase.
class Order(models.Model):
poid = models.IntegerField(db_column='POID', primary_key=True) # Field name made lowercase.
po = models.CharField(db_column='PO', unique=True, max_length=20) # Field name made lowercase.
quote = models.CharField(db_column='Quote', unique=True, max_length=20) # Field name made lowercase.
condition = models.CharField(db_column='Condition', max_length=15) # Field name made lowercase.
I have tried relate_name for foreign key but still same behaviour. Same values can be stored on database without any issues. Only Django throws an error.
Please someone help me!!!
You should paste the actual error; it is presumably that "List.pid" must be an "Order" instance.
The error should be clear. pid is a ForeignKey field, it expects an instance of the related model. You can either get the item and set it:
pid = Order.objects.get(pk=3)
List(...., pid=pid)
or use the underlying id field:
List(...., pid_id=3)
The pid column at the Django level is not an id (integer), yes on the database level it is. But in Django it is a foreign key, and thus it refers to an Order instance. You thus should pass it an Order instance (the one that corresponds to poid=3).
But we are lucky, in case you construct such ForeignKey, Django automatically makes a field_id column that stores the id, and those two fields act like twins. We can thus assign 3 to pid_id:
sav_list = List(
id=4,
item_name ='name1',
item_desc='desc1',
location='location', reason='rfp',
pid_id=3
)
Note however that for most database backend, foreign key constraints will check if there is an Order instance with as referenced column (in Django the primary key) 3 exists. If not, it is impossible to create such List. I would therefore advice to ensure that such Order exists before trying to save it to the database.
I have two models:
class Amodel(models.Model):
name = models.CharField(max_length=8)
desc = models.CharField(max_length=256)
class Bmodel(models.Model):
name = models.CharField(max_length=8)
desc = models.CharField(max_length=256)
now I have another model:
class Cmodel(models.Model):
name = models.CharField(max_length=8)
f_model = models.ForeignKey(to='there I want to dynamic refers to Amodle or Bmodel when create the Cmodel instance')
I want the Cmodel's f_model is choosable when Create the Cmodel instance, whether this is possible?
This feature called generic relations. Here is the official documentation link generic-relations
By definition of foreign key you can not assign foreign key to one field with choices of model
A FOREIGN KEY is a key used to link two tables together.
A FOREIGN KEY is a field (or collection of fields) in one table that refers to the PRIMARY KEY in another table.
Instead you can proceed to create two fields as below:
class Cmodel(models.Model):
name = models.CharField(max_length=8)
f_a_model = models.ForeignKey(Amodel, blank=True, null=True, on_delete=models.CASCADE)
f_b_model = models.ForeignKey(Bmodel, blank=True, null=True, on_delete=models.CASCADE)
This way you can create two fields and you can keep it as null.
So If you wish to proceed for Cmodel instance with foreign key of a model you can add it to field f_a_model and keep f_b_model null and vice versa
You may follow example of using generic-relations from this link and the doc.
When you use generic relations you need to write your own custom field and method for serializer or form or anywhere you wish to user it.
The Model class are generated by the django inspectdb and i can see the Id field, this is converted from the Legacy tables, the table cant be altered here,
class workFlowTemplate(models.Model):
rev = models.IntegerField(null=True, blank=True)
revtype = models.IntegerField(null=True, blank=True)
**id = models.IntegerField(null=True, blank=True)**
name = models.CharField(max_length=255, blank=True,primary_key=False)
class Meta:
db_table = 'workflow_template'
the problem is when i try to run the django it throws error
CommandError: One or more models did not validate:
pHApp.workflowtemplate: "id": You can't use "id" as a field name, because each model automatically gets an "id" field if none of the fields have primary_key=True. You need to either remove/rename your "id" field or add primary_key=True to a field.
Note
the id is not primary key here in this case,
When i tries to change
id = models.IntegerField(null=True, blank=True,primary_key=False)
the Issue still remains the same, and same error is repeated again, how to avoid this Issue ?
A model must have a primary key, so you need specify a primary key field, like this:
my_primary_key = models.AutoField(primary_key=True)
id = models.IntegerField(null=True, blank=True)
then it won’t add the automatic id column and you will not receive this error.
Column id is adding by django by default, if you did not specified another field as PK.
In your case, you specified id as your field, but you did not specified what field should be a primary key.
Solution - set another field as primary key since you can not rename id