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.
Related
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!
Basically, what I want is a field to be available if a condition is met, so something like this:
class ConditionalModel(models.Model):
product = models.ForeignKey(product, on_delete=models.CASCADE)
if category == "laptop":
cpu_model = models.CharField(max_length=200)
so if I were to go to the Django admin page and create an instance of the model and then choose "laptop" as the product from the drop-down list of existing "products", a new field would be available. I couldn't find anything about this in the documentation, so I'm wondering whether it's even possible.
What you are asking for is not "technically" possible. A model relates a database object, and under traditional SQL rules, this isn't possible. You could instead make that field optional, and then customize the admin page's functionality.
Another potential option, though I do not have much experience with it, would be to use a NoSQL database in the case where you don't want to store NULL values in your db.
I do not think it is possible because models defines databases tables so the column has to be present.
You can use the keyword blank=True to allow an object without this field.
Maybe you can customize the admin interface to hide the field in some cases.
You can't do that in models.
You can hide it in admin panel or you can make separate model for laptop.
Or you can make field blank=True
Making a field optional is not possible but you can use a generalized model called Product and two or more specialized ones called for example : ElectronicProduct that contains the field cpu_model and NonElectronicProduct, the two specialized models have to contain a OneToOneField to the Product model to ensure inheritance.
I have looked for django doc in their official site but i can't find the article about the on_update model function here in Related objects reference except for on_delete.
Here is an example code:
from django.db import models
class Reporter(models.Model):
# ...
pass
class Article(models.Model):
reporter = models.ForeignKey(Reporter, on_delete=models.CASCADE)
Is there any version of on_update?
I have visited this Cascade on update and delete wih django but there is not a clear answer about the on_update
I am using mysql and define the relationship in the ERD and sync it to the db and tried running the python manage.py inspectdb to generate the django-model but it shows only models.DO_NOTHING.
Is there a better way to achieve this, if any?
It's normally adviseable to completely leave the primary key alone when setting up your Django models, as these are used by Django in a number of ways to maintain relationships between objects. Django will set them up and use them automatically.
Instead, create a separate field in your model to keep track of unique data:
class Reporter(models.Model):
emp_id = models.CharField(unique=True)
This way you can obtain the emp_id with reporter_object.emp_id and if you need it, you can still get the pk with reporter_object.id.
You can read about how it works it in the Django 1.9 Documentation
I'm really really confused about how django handles database relationships.
Originally I had an article model that contained a simple IntegerField for article_views, recently I'm trying to expand the definition of a article_view to contain it's own fields so I created a model for it. (IP, SESSION KEY etc..)
I'm at a bit of a loss regarding how to make the relationship, to me it makes the most sense to have a one-to-many field inside the article model, because an article can have many different views, but a view can only be part of one article.
all the implementations I'm seeing have this set up in a really weird reverse manner, what gives?
Unfortunately Django does not have a One-to-Many field. This is achieved by creating a ForeignKey on in this case the ArticleView model. When you want to easily access the article views in your template you can set the related_name on the ForeignKey.
class Article(models.Model):
# Article definition
class ArticleView(models.Model):
article = models.ForeignKey(Article, related_name='views')
In the template you can now use article.views.count() to get the number of views coupled to an account.
Please note that this creates a database query for each count you want. It would probably be better to have a queryset with annotate: Article.objects.annotate(num_views=Count('views'))
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.