I have these models:
class Company(models.Model):
name=models.CharField(max_length=100)
description=models.TextField()
#some more fields
class Product(models.Model):
name=models.CharField(max_length=100)
company=models.ForeignKey(Company)
#some more fields
class Category(models.Model):
parent=models.ForeignKey('self',null=True,blank=True)
name=models.CharField(max_length=100)
products=models.ManyToManyField(Product,null=True,blank=True)
#some more fields
as U can see each company has a list of product and each product belongs to some categories,I'm going to get the list of categories of each company using company pk,what's the best practice?should I define a database view?how can I do this?
Note:I've not ever used database view in django,I searched about it and that doesn't sound easy to me!
I always try to avoid using database views, stored procedures and in general stuff that 'lives' in the database itself rather than in the application code-base for the simple reason that it is very hard to maintain (and also you say good bye to database agnostic applications).
My advice here is to stick with django orm (which can do a lot) and only if you unable to get decent performances or if you need some advanced feature available through stored procedures/views only then to go for that solution.
Using views in django is quite easy.
Say you have 1 view to query, you create the view on the db then you write the model with fields matching the view' columns (name and type).
UPDATE:
You then need to set the table name as the view name in meta class definition.
After that you need to tell django not to write on that and to not try to create a table for the view model, luckily there is a conf for that:
class ViewModel(models.Model):
... view columns ...
class Meta():
db_table = 'view_name'
managed = False
I've no idea why you think you need a db view here. Generally, you don't use them with Django, since you do all the logic in Python via the ORM.
To get the list of categories for a company, you can just do:
categories = Category.objects.filter(products__company=my_company)
where my_company is the Company instance you're interested in.
Related
I am trying to use a same table across two different projects and want to reflect the changes between them.
Let's assume that there is a student table in the data created by PROJECT1-APP1 of django. I want that same student table to use in PROJECT2-APP1. I have searched the interned and get recommendation to import the model from 1 project to other but according to my understanding it is just "faking" it and some functionality like BigAutoFeild and Foreign Keys would not work in this solution. Is there a way that I can use the same database in two different projects which will not mess up with the core operation?
EDIT:
the other questions just shows how to connect database which I know, the problem occurs when I try to use the same tables.
You may need to create a model in the other app and explicitly specify the database table in the class Meta section. Similar to below:
class Person(models.Model):
id = models.IntegerField(primary_key=True)
first_name = models.CharField(max_length=70)
class Meta:
db_table = 'table_persons'
I am trying to build a Django website where the user is able to create custom objects known as items. Each item needs to be able to have certain properties that are stored in the database. For example an item would need properties such as
Serial Number,
Description,
Manufacture Date
However I want the user to be able to specify these fields similar to what Microsoft dynamics allows . For example a user should be able to specify they want a text field with the name Model Number, associated with a specific item type and from then on they can store those properties in the database.
I am not sure the best approach to do this because a standard database model, you already have all the fields defined for a specific table, however this essentially means i have to find a way to have user defined tables.
Does anyone know a good approach to handle this problem, at the end of the day I want to store items with custom properties as defined by the user in a database.
thanks
There are multiple ways you can go.
In non-relational databases you don't need to define all the fields for a collections ( analogous to a table of RDBMS).
But if you want to use SQL with Django, then you can define a Property Model.
class Property(models.Model):
name = CharField()
value = CharField()
item = models.ForeignKey(Item, on_delete=models.CASCADE)
class Item(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
You can render a FormSet of Property form. To add extra empty forms on the fly, render dynamic formsets.
I'm new to Django and I am trying to use a mysql database created and filled with data by someone else
I created a model with the same name as the table I want to get data from, my models is as follows
class Study(models.Model):
study_name = models.TextField(default='Unknown')
description = models.TextField(default='Unknown')
language = models.TextField(default='Unknown')
number_of_years = models.IntegerField(default='0')
database connected but when I go to admin I don't see the data there
Please help me with this
A step by step solution would be:
get the name of the table containing your data, I'll call it study_table
make sure you know how the table was defined so you can match it with django model definition. Connect to the database with a MySQL client and run the following query:
DESCRIBE study_table;
based on the table name, column types and column names, define your model to match everything. Django models do a lot of automated naming so you have force the naming to make sure your model matches your database. Principles are:
Specify the table name as a meta option.
Create fields with names matching column names and field types matching column types. Taking an example from your code, the field study_name should match a column with the same name in the table study_table.
class Study(models.Model):
study_name = models.TextField(default='Unknown')
description = models.TextField(default='Unknown')
language = models.TextField(default='Unknown')
number_of_years = models.IntegerField(default=0)
class Meta:
db_table = study_table
Side note: your IntegerField has a default as a string '0'.
making sure the app (I'll call it study_app) containing your model is enabled, the database is configure properly in your django settings, try to access data from the admin shell (python manage.py shell):
>>> from study_app.models import Study
>>> Study.objects.first()
This should return an answer, if it does not, your model doesn't match the database data.
to make accessing the data easier, create an admin page as suggested by #iklinac. You can now read, edit your data through your browser.
A few suggestions you could consider:
study_name should probably be a models.CharField(max_length=255) or similar
description should be allowed to be empty models.TextField(blank=True)
language should probably be a models.CharField with a choices option.
You should create ModelAdmin instance for your model
The ModelAdmin class is the representation of a model in the admin
interface. Usually, these are stored in a file named admin.py in your
application.
from django.contrib import admin
from myproject.myapp.models import Study
class StudyAdmin(admin.ModelAdmin):
pass
admin.site.register(Study, StudyAdmin)
If you have a MySQL database with tables of data that don't have models created yet, you can use the dumpdata command to automatically generate the models:
https://docs.djangoproject.com/en/3.0/ref/django-admin/#dumpdata
Then, you can register those models in the Django admin. dumpdata should only be used as a starting point, since they are auto-generated and won't contain many of Django's data integrity features.
Good luck!
Say we're building a Django-based site that clones Medium.com's URL structure, where you have users and articles. We'd probably have this model:
class Article(models.Model):
user = models.ForeignKey(User)
slug = models.CharField()
We want to be able to build URLs that look like /<username>/<slug>/. Since we're going to have billions of articles and zillions of pageviews, we want to put an index on that model:
class Meta:
indexes = [
models.Index(fields=['user__username', 'slug'])
]
But this causes the makemigrations command to fail with the following error:
django.core.exceptions.FieldDoesNotExist: Article has no field named 'user__username'. The app cache isn't ready yet, so if this is an auto-created related field, it won't be available yet.
So, plain vanilla models.Index doesn't support relational lookups like a QuerySet does. How would I add an index like this? Let's assume PostgreSQL, if that's helpful.
It seems that you can't make multi-table index according to this answer.
So if it's not possible in the database, I don't see how can Django offer this feature...
What you can do to make your queries more efficients is an index using user_id and slug.
Django index meta class mainly provide declarative options for indexing table fields,
you can create an index using several field of a model or create different index for every fields of the model. you just don't have to provide user foriegnkey field name attribute which generate automatic user_id index migrations
migrations.AddIndex(
model_name='candidates',
index=models.Index(fields=['user'], name='candidates__user_id_569874_idx'),
),
you can also set the index name in the model meta, and db_tablspace as well if needed.
Note: I've since asked this question again given the updates to Django's user model since version 1.5.
I'm rebuilding and making improvements to an already existing Django site and moving it over from Webfaction to Heroku, and from Amazon's SimpleDB to Heroku Postgres (though testing locally on Sqllite3 when developing). A lot of what I'm doing is moving over to use built-in Django functionality, like the Django admin, user authentication, etc.
Conceptually, the site has two kinds of users: Students and Businesses. The two types of users have completely different permissions and information stored about them. This is so much the case that in the original structure of the site, we set up the data model as follows:
Users
ID (primary_key)
Business_or_Student ('B' if business, 'S' if student)
email (unique)
password (hashed, obviously)
...
Students
ID (Foreignkey on Users)
<more information>
...
Businesses
ID (Foreignkey on Users)
<more information>
...
This worked pretty well for us, and we had the bare-bones user information in the Users table, and then any more detailed information in the Students and Businesses tables. Getting a user's full profile required something along this pseudocode:
def get_user_profile(id):
if Users(id=id).Business_or_Student = 'B':
return Businesses(id=id)
else:
return Students(id=id)
In moving over, I've found that Django's built-in User object has pretty limited functionality, and I've had to extend it with a UserProfile class I've created, and then had additional Student and Business tables. Given all of the patching I'm doing with this in the Django admin, and being relatively unfamiliar with Django models since I always did it differently, I'm not sure if this is the best way to go about it, or if I should just stick all of the information for businesses and students in the UserProfile table and just differentiate the two with different groups, or if there's even some way to do this all in the built-in User object.
Since businesses and students also have different interfaces, I'm seriously considering setting up the two as different apps within my Django project, and so separating their views, models, etc. entirely. That would look something like:
MyProject/
MyProject/ (project folder, Django 1.4)
mainsite/
students/
businesses/
One of my biggest concerns is with the Django Admin. In extending User, I already had to add the following code:
class UserProfileInline(admin.StackedInline):
model = UserProfile
can_delete = False
verbose_name_plural = 'profile'
class UserAdmin(UserAdmin):
inlines = (UserProfileInline, )
However, I would like the information for the Business or Student aspects of the user to show up in the Django admin when that User is pulled up, but the ForeignKey part of the model is in the Student and Business model since every Student/Business has a User but every User has only one Student or one Business object connected with it. I'm not sure how to add a conditional Inline for the Admin.
Question: Given this structure and these concerns, what is the best way to set up this site, particularly the data model?
This is not a complete solution, but it will give you an idea of where to start.
Create a UserProfile model in mainsite. This will hold any common attributes for both types of users. Relate it to the User model with a OneToOne(...) field.
Create two more models in each app, (student/business), Business and Student, which have OneToOne relationships each with UserProfile (or inherit from UserProfile). This will hold attributes specific to that type of users. Docs: Multitable inheritance / OneToOne Relationships
You may add a field in UserProfile to distinguish whether it is a business or student's profile.
Then, for content management:
Define the save() functions to automatically check for conflicts (e.g. There is an entry for both Business and Student associated with a UserProfile, or no entries).
Define the __unicode__() representations where necessary.
I hope I understood your problem... maybe this can work? You create a abstract CommonInfo class that is inherited in into the different Sub-classes (student and businesses)
class CommonUser(models.Model):
user = models.OneToOne(User)
<any other common fields>
class Meta:
abstract = True
class Student(CommonUser):
<whatever>
class Business(CommonUser):
<whatever>
In this case the models will be created in the DB with the base class fields in each table. Thus when you are working with Students you run a
students = Students.objects.get.all()
to get all your students including the common information.
Then for each student you do:
for student in students:
print student.user.username
The same goes for Business objects.
To get the student using a user:
student = Student.objects.get(user=id)
The username will be unique thus when creating a new Student or Business it will raise an exception if an existing username is being saved.
Forgot to add the link