I'm making one of my first django apps with sqlite database. I have some models like for example:
class Connection(models.Model):
routeID = models.ForeignKey(Route, on_delete=models.CASCADE)
activityStatus = models.BooleanField()
car = models.ForeignKey(Car, on_delete=models.CASCADE)
class Route(models.Model):
name = models.CharField(max_length=20)
and forms
class RouteForm(ModelForm):
class Meta:
model = Route
fields = ['name']
class ConnectionForm(ModelForm):
class Meta:
model = Connection
fields = ['routeID', 'activityStatus', 'car']
And in my website, in the url for adding new Connection, I have cascade list containing RouteIDs. And I'd like it to contain RouteName, not ID, so it would be easier to choose. How should I change my ConnectionForm, so I could still use foreign key to Route table, but see RouteName instead of RouteID?
For now it's looking like this, but I'd love to have list of RouteNames, while still adding to Connection table good foreign key, RouteID
Update the Route Model's __str__ method:
class Route(models.Model):
name = models.CharField(max_length=20)
def __str__(self):
return self.name
Because the __str__() method is called whenever you call str() on an object. Django uses str(obj) in a number of places like in Modelform. By default it returns id or pk that is why you were seeing ids in model form. So by overriding it with name, you will see the names appear in choice field. Please see the documentation for more details on this.
Related
It is possible to define a __str__ method on a model:
class Person(models.Model):
name = models.CharField(max_length=40)
hobbies = models.ManyToManyField(Hobby)
def __str__(self):
return self.name
But how to define something like this for an automatically generated table through a ManyToManyField? Objects that represent a relationship between a person and a hobby are written like this:
Person_hobbies object (1)
Note that I am not talking about the Hobby object itself, but the object that represents the relation between a Person and a Hobby.
But how to define something like this for an automatically generated table through a ManyToManyField?
You can define the junction table model with the through=… parameter [Django-doc]:
class Person(models.Model):
name = models.CharField(max_length=40)
hobbies = models.ManyToManyField(Hobby, through='PersonHobby')
def __str__(self):
return self.name
class PersonHobby(models.Model):
person = models.ForeignKey(Person, on_delete=models.CASCADE)
hobby = models.ForeignKey(Hobby, on_delete=models.CASCADE)
def __str__(self):
return f'A string with the {self.person} and the {self.hobby}'
Migrating to a junction model for a ManyToManyField is however not that easy. If it is possible, I would delete migrations that include the ManyToManyField and recreate a migration.
Context: I'm forcing my self to learn django, I already wrote a small php based website, so I'm basically porting over the pages and functions to learn how django works.
I have 2 models
from django.db import models
class Site(models.Model):
name = models.CharField(max_length=50, unique=True)
def __str__(self):
return self.name
class Combo(models.Model):
username = models.CharField(max_length=50)
password = models.CharField(max_length=50)
dead = models.BooleanField(default=False)
timestamp = models.DateTimeField(auto_now_add=True)
siteID = models.ForeignKey(Site, on_delete=models.PROTECT)
class Meta:
unique_together = ('username','siteID')
def __str__(self):
return f"{self.username}:{self.password}#{self.siteID.name}"
When creating a view, I want to get the Combo objects, but I want to sort them first by site name, then username.
I tried to create the view, but get errors about what fields I can order by Cannot resolve keyword 'Site' into field. Choices are: dead, id, password, siteID, siteID_id, timestamp, username
def current(request):
current = Combo.objects.filter(dead=False).order_by('Site__name','username')
return render(request, 'passwords/current.html',{'current':current})
Since I'm not necissarily entering the sites into the database in alphabetical order, ordering by siteID wouldn't be useful. Looking for some help to figure out how to return back the list of Combo objects ordered by the Site name object then the username.
You can order this by siteID__name:
def current(request):
current = Combo.objects.filter(dead=False).order_by('siteID__name','username')
return render(request, 'passwords/current.html',{'current':current})
since that is the name of the ForeignKey. But that being said, normally ForeignKeys are not given names that end with an ID, since Django already adds an _id suffix at the end for the database field.
Normally one uses:
class Combo(models.Model):
# …
site = models.ForeignKey(Site, on_delete=models.PROTECT)
if you want to give the database column a different name, you can specify that with the db_column=… parameter [Django-doc]:
class Combo(models.Model):
# …
site = models.ForeignKey(
Site,
on_delete=models.PROTECT,
db_column='siteID'
)
Looking at graphene_django, I see they have a bunch of resolvers picking up django model fields mapping them to graphene types.
I have a subclass of JSONField I'd also like to be picked up.
:
# models
class Recipe(models.Model):
name = models.CharField(max_length=100)
instructions = models.TextField()
ingredients = models.ManyToManyField(
Ingredient, related_name='recipes'
)
custom_field = JSONFieldSubclass(....)
# schema
class RecipeType(DjangoObjectType):
class Meta:
model = Recipe
custom_field = ???
I know I could write a separate field and resolver pair for a Query, but I'd prefer it to be available as part of the schema for that model.
What I realize I could do:
class RecipeQuery:
custom_field = graphene.JSONString(id=graphene.ID(required=True))
def resolve_custom_field(self, info, **kwargs):
id = kwargs.get('id')
instance = get_item_by_id(id)
return instance.custom_field.to_json()
But -- this means a separate round trip, to get the id then get the custom_field for that item, right?
Is there a way I could have it seen as part of the RecipeType schema?
Ok, I can get it working by using:
# schema
class RecipeType(DjangoObjectType):
class Meta:
model = Recipe
custom_field = graphene.JSONString(resolver=lambda my_obj, resolve_obj: my_obj.custom_field.to_json())
(the custom_field has a to_json method)
I figured it out without deeply figuring out what is happening in this map between graphene types and the django model field types.
It's based on this:
https://docs.graphene-python.org/en/latest/types/objecttypes/#resolvers
Same function name, but parameterized differently.
Probably a very novice Django question, but here goes. In my Django project, I have this in my models
#models.py
class Classes(models.Model):
classcode = models.CharField(max_length=15)
classname = models.TextField()
students = models.ManyToManyField(User)
class Test(models.Model):
classes = models.ForeignKey(Classes, on_delete=models.CASCADE)
name = models.TextField(max_length=100)
points = models.ManyToManyField(User, default=0)
I also have a form for Test, which is:
#forms.py
class TestForm(forms.ModelForm):
class Meta:
model = Test
fields = ('classes', 'name')
When I get to the actual form, the drop-down menu for 'classes' in TestForm merely comes up with 'Classes object' for the number of 'Classes' that I have in my DB. I want to change that so the form lists the names of the classes, which are stored in the 'Classes' model as 'classname'
Can anyone point me in the right direction please?
The easiest way to do it is to provide a string representation of your object, this would replace any where you access the class throughout your application
class Classes(models.Model):
classcode = models.CharField(max_length=15)
classname = models.TextField()
students = models.ManyToManyField(User)
def __str__(self):
return "{0}: {1}".format(self.classcode, self.classname)
From the docs
The __str__ (__unicode__ on Python 2) method of the model will be called to generate string representations of the objects for use in the field’s choices; to provide customized representations, subclass ModelChoiceField and override label_from_instance.
I have the following code:
class Item(models.Model):
name = models.CharField(max_length=100)
keywords = models.CharField(max_length=255)
type = models.ForeignKey(Type)
class Meta:
abstract = True
class Variant(models.Model):
test_field = models.CharField(max_length=255)
class Product(Item):
price = models.DecimalField(decimal_places=2, max_digits=8,null=True, blank=True)
brand = models.ForeignKey(Brand)
variant = models.ForeignKey(Variant)
def get_fields(self):
return [(field.name, field.value_to_string(self)) for field in Product._meta.fields]
def __unicode__(self):
return self.name
Im using Grappelli.
I want my Product to have multiple Variations. Should I use a manytomanyfield?
I want to be able to add Variants to my Product directly in the Admin. Now I get an empty dropwdown with no variants(because they doesnt exists).
I thought Django did this automatically when u specified a Foreign Key?
How can I get the Variant fields to display directly on my Product page in edit?
I've read someting about inline fields in Admin?
Well, it's the other way around :)
1/ Place the foreign key field in your Variant, not in your Product (what you describe is actually a OneToMany relationship).
2/ Link the Variant to your Product in the relative ProductAdmin in admin.py as an inline (i.e VariantInline).
See the docs for further informations : https://docs.djangoproject.com/en/1.6/ref/contrib/admin/#inlinemodeladmin-objects
Hope this helps !
Regards,