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
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 creating a Django web app. I have set a field as a primary key and a unique field. When I try to edit it and give it the same value as another item in the primary key column, the current item got duplicated and each value goes to one of them as shown below.
This is my code from models.py:
class Car(models.Model):
class Meta:
verbose_name = "Car"
verbose_name_plural = "Cars"
car_id = models.CharField(verbose_name='Car ID', max_length=10000, primary_key=True, unique=True)
Some screenshots from the admin panel:
Once you change the primary key, Django will execute an update statement for that id so corrputing your data, the best way as mentioned by #vishal-singh, is to set the id field as Read-only.
I'm trying to query through a model in Django that has no Primary Key.
I need to query though it so I can access to the Foreign keys it has.
I'm just trying to do this atm and doesn't even work:
chars = Characterweapons.objects.all()
print(chars)
And if i change Characterweapons to Weapons for example, a table with Primary Key it works.
The error I get when I load the page is this:
Exception Value: (1054, "Unknown column 'characterweapons.id' in 'field list'")
This is my model:
class Characterweapons(models.Model):
characterid = models.ForeignKey(Characters, models.DO_NOTHING, db_column='CharacterID') # Field name made lowercase.
weaponid = models.ForeignKey(Weapons, models.DO_NOTHING, db_column='WeaponID', blank=True, null=True) # Field name made lowercase.
categoryid = models.ForeignKey(Category, models.DO_NOTHING, db_column='CategoryID', blank=True, null=True) # Field name made lowercase.
quantity = models.IntegerField(db_column='Quantity', blank=True, null=True) # Field name made lowercase.
class Meta:
managed = False
db_table = 'characterweapons'
def __str__(self):
return '%s %s %s %s' % (self.quantity,self.characterid,self.weaponid,self.categoryid)
Anyone knows about this?
Thanks in advance!
So it looks like you did a DBinspect on an existing database to generate this model. I'm guessing this is failing because Django ORM expects your table to have a primary key. "id" is the default name for a Django generated model primary key field. I suspect when you are trying to call Characterweapons.objects.all() it's trying to get the primary key field for some reason.
There may be a workaround but unless you really know what your doing with your database, I would highly urge you to set a primary key on your tables.
The best solution that I found, in this case, was to perform my own query, for example:
fact = MyModelWithoutPK.objects.raw("SELECT * FROM my_model_without_pk WHERE my_search=some_search")
This way you don't have to implement or add another middleware.
see more at Django docs
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 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)