Develop Reference

Using Relationship Based Data in Model Definition - Django

I want to make the string representation of a field show data based on a JOIN, for instance:
For Penciler - I want the string representation to resolve to
John Doe (DC) - But the publisher value in that class is a Foreign Key - How do I reference the publisherName?
from django.db import models
class Series(models.Model):
seriesId = models.AutoField(primary_key=True)
series_name = models.CharField(max_length=300)
publisher = models.ForeignKey('Publisher', on_delete = models.PROTECT)
first_published = models.DateField()
last_published = models.DateField()
discontinued = models.BooleanField(default=False)
def __str__(self):
return f'{self.series_name} - {self.publisher} ({self.first_published - self.last_published})'
class Meta:
ordering = ['publication_year','title']
class Publisher(models.Model):
publisherId = models.AutoField(primary_key=True)
publisherName = models.CharField(max_length=250, null=False)
def __self__(self):
return self.publisherName
class Penciler(models.Model):
pencilerID = models.AutoField(primary_key=True)
pencilerName = models.CharField(max_length=200)
publisher = models.ForeignKey('Publisher', on_delete= models.PROTECT)
def __str__(self):
return self.pencilerName (self.publisher)

You can access the related Publisher instance through the ForeignKey field and get the publisherName in the __str__() method so:
def __str__(self):
publisher_name = self.publisher.publisherName
return f'{self.pencilerName} ({publisher_name})'
Additionally, I'd recommend you to use string formatting, such as using f-strings.

It is as simple as that:
def __str__(self):
return f"{self.pencilerName} ({self.publisher})"

Django- Can only concatenate str (not "ManyRelatedManager") to str

I am trying to save my model. But when I try to save my model I throws the following error
TypeError at /admin/user/teacher/add/
can only concatenate str (not "ManyRelatedManager") to str
My models.py file looks like this
class Class(models.Model):
Class = models.CharField(max_length=50)
section_choices = (('A','A'),('B','B'),('C','C'),('D','D'),('E','E'))
Section = models.CharField(max_length=100, choices=section_choices)
def __str__(self):
return self.Class + "," + self.Section
class Subject(models.Model):
subject = models.CharField(max_length=100)
def __str__(self):
return self.subject
class Teacher(models.Model):
User = models.ForeignKey(User, on_delete=models.CASCADE)
Subject = models.ManyToManyField(Subject)
Name = models.CharField(max_length=50)
Profile = models.ImageField(upload_to = upload_teacher_profile_to, default =
'defaults/teacher_profile.png')
Class = models.ManyToManyField(Class)
Number = models.IntegerField(blank=True)
is_banned = models.BooleanField(default=False)
def __str__(self):
return self.Name + "of" + self.Class

Updated your Teacher Model def __str__(self) function to
def __str__(self):
return f"{self.Name} of {[class for class in self.Class.all()]}"
Class is a ManyToManyField in the Teacher Model, ManyToManyFields can contain many objects, hence you cannot use self.Class
You can learn more about ManyToManyField here

Try with(in class Teacher):
def __str__(self):
output=""
for x in self.Class.all():
output=output+' | '+str(x)
return output

How to show foreignkey attributes django admin fields?

This question is similar with others but it is a different one actually ! So, I have 3 models such as (I have deleted some unnecessary things for shorter code):
class Category(models.Model):
category_title = models.CharField(max_length=200)
category_content = models.TextField()
category_slug = models.CharField(max_length=200)
def __str__(self):
return self.category_title
class Classes(models.Model):
classes_title = models.CharField(max_length=200)
classes_content = models.TextField()
classes_category = models.ForeignKey(Category, on_delete=models.SET_DEFAULT)
def __str__(self):
return self.classes_title
class Subjects(models.Model):
subject_title = models.CharField(max_length=200)
subject_content = models.TextField()
subject_class = models.ForeignKey(Classes, on_delete=models.SET_DEFAULT)
def __str__(self):
return self.subject_title
So let me give an example. I can have 2 categories and in those categories I can have "same named" classes. Lets think about maths is a class for both categories. When I want to add a new subject to maths I see 2 same named maths in admin page. So I want to know which one belongs to which category in admin page. So I can add my subject to right class.
class SubjectAdmin(admin.ModelAdmin):
fields = ('subject_title', 'subject_content', 'subject_class',)
So in this picture (Subjects = Konular) I am adding a new subject. I will have a foreign key to Class. However I have same name for classes that are coming from different categories. So in this dropdown how can I know which class belongs to which category ?

Try this...
class KonularAdmin(admin.ModelAdmin):
fields = ('subject_title', 'subject_content', 'subject_class', 'get_classes_category_title')
def get_classes_category_title(self, obj):
subject_object = Subjects.objects.get(id=obj.subject_class)
return str(subject_object.classes_category.category_title)
It returns the category title name

If I understood you correctly, This should work.
class Classes(models.Model):
classes_title = models.CharField(max_length=200)
classes_content = models.TextField()
classes_category = models.ForeignKey(Category, on_delete=models.SET_DEFAULT)
def __str__(self):
return self.title
class Subjects(models.Model):
subject_title = models.CharField(max_length=200)
subject_content = models.TextField()
subject_class = models.ForeignKey(Classes, on_delete=models.SET_DEFAULT)
def __str__(self):
return f"{self.subject_title} - {str(self.subject_class)}"

You can use __str__() method to change the string representation of an object:
class Subjects(models.Model):
subject_title = models.CharField(max_length=200)
subject_content = models.TextField()
subject_class = models.ForeignKey(Classes, on_delete=models.SET_DEFAULT)
def __str__(self):
return f"{self.subject_title} - {self.subject_content} - {self.subject_class}"
# shorter version:
# return f"{self.subject_title[:10]} - {self.subject_content[:10]} - {self.subject_class[:10]}"
Check it with:
>>> print(Subjects.objects.first())

ValueError: Cannot assign "<Truckdb: Truckdb object (1)>": "Quiz.truck_name" must be a "truck_name" instance

I am trying to create an instance in my app like this:
Views.py
new_quiz = Quiz.objects.create(owner=request.user, comments="Autogenerated", truck_type=truck_type_object,
truck_name=chosen_truck_object)
where chosen_truck_object is this:
chosen_truck_object = Truckdb.objects.filter(display_name=chosentruck)[0]
And Models.py
class Quiz(models.Model):
owner = models.ForeignKey(User, on_delete=models.CASCADE, related_name='quizzes')
comments = models.TextField(max_length=256, blank=True)
truck_type = models.ForeignKey(truck_type, on_delete=models.CASCADE, related_name='trucks')
truck_name = models.ForeignKey(truck_name, on_delete=models.SET_NULL, null=True)
class truck_type(models.Model):
name = models.CharField(max_length=30)
color = models.CharField(max_length=7, default='#007bff')
def __str__ (self):
return self.name
class truck_name(models.Model):
truck_type = models.ForeignKey(truck_type, on_delete=models.CASCADE)
name = models.CharField(max_length=30)
def __str__ (self):
return self.name
How can I pass the truck_type and truck_name instance to the Quiz model in Quiz.objects.create ?

Firstly you need to follow the naming convention guidelines, so your models' name must be camelcase doc as like:
class TruckType(models.Model):
name = models.CharField(max_length=30)
color = models.CharField(max_length=7, default='#007bff')
def __str__ (self):
return self.name
class TruckName(models.Model):
truck_type = models.ForeignKey(TruckType, on_delete=models.CASCADE)
name = models.CharField(max_length=30)
def __str__ (self):
return self.name
And then please migrate your database and then for your problem you need to take TruckName object instead of Truckdb.

chosen_truck_object = TruckName.objects.filter(display_name=chosentruck)[0]
instead of filter use get method chosen_truck_object = TruckName.objects.get(display_name=chosentruck) it will save

Invalid literal for int() with base 10: 'marlm'

I have to create a view which returns me the posts of a particular owner/author
This error in coming from view bellow. I know this is a simple thing that I am missing. Please help me on this: (marlm is the ID I created to test)
def view_by_owner(request):
user = request.user.username
posts_owner = Post.objects.filter(owner=user)
return render_to_response('view_post.html',{'view_owner':posts_owner})
Models:
from django.db import models
from django.db.models import permalink
from django.contrib.auth.models import User
class Post(models.Model):
title = models.CharField(max_length=100)
body = models.TextField()
datposted = models.DateTimeField('date posted')
category = models.ForeignKey('Category')
owner = models.ForeignKey('UserProfile')
def __str__(self):
return '%s' % self.title
class Category(models.Model):
title = models.CharField(max_length=100)
def __str__(self):
return self.title
class UserProfile(models.Model):
# This line is required. Links UserProfile to a User model instance.
user = models.OneToOneField(User)
# The additional attributes we wish to include.
website = models.URLField(blank=True)
picture = models.ImageField(upload_to='profile_images', null=True)
# Override the __unicode__() method to return out something
def __unicode__(self):
return self.user.username
class Logout(User):
force_logout_date = models.DateTimeField(null=True, blank=True)

The problem is that owner is a foreign key to User, not the value of the user name. To lookup by that, you will need to span that relationship, like so:
posts_owner = Post.objects.filter(owner__user__username=user)
Or, more simply, you could do this:
posts_owner = Post.objects.filter(owner__user=request.user)