I have the following models:
class Model1(models.Model):
...
class Model2(models.Model):
...
model1 = models.ForeignKey(Model1)
Now, lets say I have an object of Model2 with pk=241 which is related to another object of Model1 with pk=102. I am querying them as follows:
model2 = Model2.objects.get(pk=241)
Now, if I want the pk of the referenced Model1 object. I do the following:
model2.model1.pk
This should not query the database again according to what I understand about tables, but if I run the following:
from django.db import connection
connection.queries
I get a list of 2 queries. Why do I need to query my database again to only get the primary key of my related object? Is there a way to avoid doing this?
I am aware of select_related(), however, what if I want to call the Model1 objects pk in the save() method of the Model2 class?
Moreover, is select_related() required even if I want to just retrieve the pk of the related object and nothing more?
You can access the underlying field without a db hit.
model2.model1_id
You don't need select_related here, since you are not actually accessing the related object.
I have following model and extracted queryset using prefetch_related as below.
queryset = Light.objects.filter(
certificate__name="A").prefetch_related('zone__namingzone'
)
From this queryset, I want to get following data set.
{"naming1":lpd1,"naming2":lpd2...}
However, when I try to extract attribute from queryset as below, I get create_reverse_many_to_one_manager
for i in queryset:
print (i.zone.namingzone)
What I want to get is naming attribute in naming table. Could anyone tell me how I can extract this?
models.py
class Certificate(models.Model):
name=models.CharField(max_length=20)
class Zone(models.Model):
zone=models.CharField(max_length=20)
class Light(models.Model):
certificate=models.ForeignKey(Certificate, on_delete=models.CASCADE,related_name='certificate')
zone=models.ForeignKey(Zone, on_delete=models.CASCADE,related_name='lightzone')
lpd=models.IntegerField()
class Meta:
unique_together = (('certificate', 'zone'),)
class Naming(models.Model):
zone=models.ForeignKey(Zone, on_delete=models.CASCADE,related_name='namingzone')
naming=models.CharField(max_length=20)
When you traverse a FK in reverse, you end up with a manager, and multiple items on the other side. So i.zone.namingzone in your for loop is a manager, not a NamingZone. If you change your print loop to:
for i in queryset:
print (i.zone.namingzone.all())
You should see all the naming zones for your item. You can extract the naming field from each NamingZone from the queryset as follows:
queryset.values('zone__namingzone__naming')
You probably want to extract a few other fields from your Light model, like the lpd for instance:
queryset.values('lpd', 'zone__namingzone__naming')
You might have a same ldp several times, as many times as it has naming zones.
Consider the following many-to-one relationship. Users can own many widgets:
class Widget(models.Model):
owner = models.ForeignKey('User')
class User(models.Model):
pass
I have a fairly complicated query that returns me a queryset of Widgets. From that I want to list the distinct values of User.
I know I could loop my Widgets and pull out users with .values('owner') or {% regroup ... %} but the dataset is huge and I'd like to do this at the database so it actually returns Users not Widgets first time around.
My best idea to date is pulling out a .values_list('owner', flat=True) and then doing a User.objects.filter(pk__in=...) using that. That pulls it back to two queries but that still seems like more than it should need to be.
Any ideas?
Use backward relationship:
User.objects.distinct().filter(widget__in=your_widget_queryset)
I'm trying to optimise my app by keeping the number of queries to a minimum... I've noticed I'm getting a lot of extra queries when doing something like this:
class Category(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=127, blank=False)
class Project(models.Model):
categories = models.ManyToMany(Category)
Then later, if I want to retrieve a project and all related categories, I have to do something like this :
{% for category in project.categories.all() %}
Whilst this does what I want it does so in two queries. I was wondering if there was a way of joining the M2M field so I could get the results I need with just one query? I tried this:
def category_list(self):
return self.join(list(self.category))
But it's not working.
Thanks!
Which, whilst does what I want, adds an extra query.
What do you mean by this? Do you want to pick up a Project and its categories using one query?
If you did mean this, then unfortunately there is no mechanism at present to do this without resorting to a custom SQL query. The select_related() mechanism used for foreign keys won't work here either. There is (was?) a Django ticket open for this but it has been closed as "wontfix" by the Django developers.
What you want is not seem to possible because,
In DBMS level, ManyToMany relatin is not possible, so an intermediate table is needed to join tables with ManyToMany relation.
On Django level, for your model definition, django creates an ectra table to create a ManyToMany connection, table is named using your two tables, in this example it will be something like *[app_name]_product_category*, and contains foreignkeys for your two database table.
So, you can not even acces to a field on the table with a manytomany connection via django with a such categories__name relation in your Model filter or get functions.
This is a problem concerning django.
I have a model say "Automobiles". This will have some basic fields like "Color","Vehicle Owner Name", "Vehicle Cost".
I want to provide a form where the user can add extra fields depending on the automobile that he is adding. For example, if the user is adding a "Car", he will extra fields in the form, dynamically at run time, like "Car Milage", "Cal Manufacturer".
Suppose if the user wants to add a "Truck", he will add "Load that can be carried", "Permit" etc.
How do I achieve this in django?
There are two questions here:
How to provide a form where the user can add new fields at run time?
How to add the fields to the database so that it can be retrieved/queried later?
There are a few approaches:
key/value model (easy, well supported)
JSON data in a TextField (easy, flexible, can't search/index easily)
Dynamic model definition (not so easy, many hidden problems)
It sounds like you want the last one, but I'm not sure it's the best for you. Django is very easy to change/update, if system admins want extra fields, just add them for them and use south to migrate. I don't like generic key/value database schemas, the whole point of a powerful framework like Django is that you can easily write and rewrite custom schemas without resorting to generic approaches.
If you must allow site users/administrators to directly define their data, I'm sure others will show you how to do the first two approaches above. The third approach is what you were asking for, and a bit more crazy, I'll show you how to do. I don't recommend using it in almost all cases, but sometimes it's appropriate.
Dynamic models
Once you know what to do, this is relatively straightforward. You'll need:
1 or 2 models to store the names and types of the fields
(optional) An abstract model to define common functionality for your (subclassed) dynamic models
A function to build (or rebuild) the dynamic model when needed
Code to build or update the database tables when fields are added/removed/renamed
1. Storing the model definition
This is up to you. I imagine you'll have a model CustomCarModel and CustomField to let the user/admin define and store the names and types of the fields you want. You don't have to mirror Django fields directly, you can make your own types that the user may understand better.
Use a forms.ModelForm with inline formsets to let the user build their custom class.
2. Abstract model
Again, this is straightforward, just create a base model with the common fields/methods for all your dynamic models. Make this model abstract.
3. Build a dynamic model
Define a function that takes the required information (maybe an instance of your class from #1) and produces a model class. This is a basic example:
from django.db.models.loading import cache
from django.db import models
def get_custom_car_model(car_model_definition):
""" Create a custom (dynamic) model class based on the given definition.
"""
# What's the name of your app?
_app_label = 'myapp'
# you need to come up with a unique table name
_db_table = 'dynamic_car_%d' % car_model_definition.pk
# you need to come up with a unique model name (used in model caching)
_model_name = "DynamicCar%d" % car_model_definition.pk
# Remove any exist model definition from Django's cache
try:
del cache.app_models[_app_label][_model_name.lower()]
except KeyError:
pass
# We'll build the class attributes here
attrs = {}
# Store a link to the definition for convenience
attrs['car_model_definition'] = car_model_definition
# Create the relevant meta information
class Meta:
app_label = _app_label
db_table = _db_table
managed = False
verbose_name = 'Dynamic Car %s' % car_model_definition
verbose_name_plural = 'Dynamic Cars for %s' % car_model_definition
ordering = ('my_field',)
attrs['__module__'] = 'path.to.your.apps.module'
attrs['Meta'] = Meta
# All of that was just getting the class ready, here is the magic
# Build your model by adding django database Field subclasses to the attrs dict
# What this looks like depends on how you store the users's definitions
# For now, I'll just make them all CharFields
for field in car_model_definition.fields.all():
attrs[field.name] = models.CharField(max_length=50, db_index=True)
# Create the new model class
model_class = type(_model_name, (CustomCarModelBase,), attrs)
return model_class
4. Code to update the database tables
The code above will generate a dynamic model for you, but won't create the database tables. I recommend using South for table manipulation. Here are a couple of functions, which you can connect to pre/post-save signals:
import logging
from south.db import db
from django.db import connection
def create_db_table(model_class):
""" Takes a Django model class and create a database table, if necessary.
"""
table_name = model_class._meta.db_table
if (connection.introspection.table_name_converter(table_name)
not in connection.introspection.table_names()):
fields = [(f.name, f) for f in model_class._meta.fields]
db.create_table(table_name, fields)
logging.debug("Creating table '%s'" % table_name)
def add_necessary_db_columns(model_class):
""" Creates new table or relevant columns as necessary based on the model_class.
No columns or data are renamed or removed.
XXX: May need tweaking if db_column != field.name
"""
# Create table if missing
create_db_table(model_class)
# Add field columns if missing
table_name = model_class._meta.db_table
fields = [(f.column, f) for f in model_class._meta.fields]
db_column_names = [row[0] for row in connection.introspection.get_table_description(connection.cursor(), table_name)]
for column_name, field in fields:
if column_name not in db_column_names:
logging.debug("Adding field '%s' to table '%s'" % (column_name, table_name))
db.add_column(table_name, column_name, field)
And there you have it! You can call get_custom_car_model() to deliver a django model, which you can use to do normal django queries:
CarModel = get_custom_car_model(my_definition)
CarModel.objects.all()
Problems
Your models are hidden from Django until the code creating them is run. You can however run get_custom_car_model for every instance of your definitions in the class_prepared signal for your definition model.
ForeignKeys/ManyToManyFields may not work (I haven't tried)
You will want to use Django's model cache so you don't have to run queries and create the model every time you want to use this. I've left this out above for simplicity
You can get your dynamic models into the admin, but you'll need to dynamically create the admin class as well, and register/reregister/unregister appropriately using signals.
Overview
If you're fine with the added complication and problems, enjoy! One it's running, it works exactly as expected thanks to Django and Python's flexibility. You can feed your model into Django's ModelForm to let the user edit their instances, and perform queries using the database's fields directly. If there is anything you don't understand in the above, you're probably best off not taking this approach (I've intentionally not explained what some of the concepts are for beginners). Keep it Simple!
I really don't think many people need this, but I have used it myself, where we had lots of data in the tables and really, really needed to let the users customise the columns, which changed rarely.
Database
Consider your database design once more.
You should think in terms of how those objects that you want to represent relate to each other in the real world and then try to generalize those relations as much as you can, (so instead of saying each truck has a permit, you say each vehicle has an attribute which can be either a permit, load amount or whatever).
So lets try it:
If you say you have a vehicle and each vehicle can have many user specified attributes consider the following models:
class Attribute(models.Model):
type = models.CharField()
value = models.CharField()
class Vehicle(models.Model):
attribute = models.ManyToMany(Attribute)
As noted before, this is a general idea which enables you to add as much attributes to each vehicle as you want.
If you want specific set of attributes to be available to the user you can use choices in the Attribute.type field.
ATTRIBUTE_CHOICES = (
(1, 'Permit'),
(2, 'Manufacturer'),
)
class Attribute(models.Model):
type = models.CharField(max_length=1, choices=ATTRIBUTE_CHOICES)
value = models.CharField()
Now, perhaps you would want each vehicle sort to have it's own set of available attributes. This can be done by adding yet another model and set foreign key relations from both Vehicle and Attribute models to it.
class VehicleType(models.Model):
name = models.CharField()
class Attribute(models.Model):
vehicle_type = models.ForeigngKey(VehicleType)
type = models.CharField()
value = models.CharField()
class Vehicle(models.Model):
vehicle_type = models.ForeigngKey(VehicleType)
attribute = models.ManyToMany(Attribute)
This way you have a clear picture of how each attribute relates to some vehicle.
Forms
Basically, with this database design, you would require two forms for adding objects into the database. Specifically a model form for a vehicle and a model formset for attributes. You could use jQuery to dynamically add more items on the Attribute formset.
Note
You could also separate Attribute class to AttributeType and AttributeValue so you don't have redundant attribute types stored in your database or if you want to limit the attribute choices for the user but keep the ability to add more types with Django admin site.
To be totally cool, you could use autocomplete on your form to suggest existing attribute types to the user.
Hint: learn more about database normalization.
Other solutions
As suggested in the previous answer by Stuart Marsh
On the other hand you could hard code your models for each vehicle type so that each vehicle type is represented by the subclass of the base vehicle and each subclass can have its own specific attributes but that solutions is not very flexible (if you require flexibility).
You could also keep JSON representation of additional object attributes in one database field but I am not sure this would be helpfull when querying attributes.
Here is my simple test in django shell- I just typed in and it seems work fine-
In [25]: attributes = {
"__module__": "lekhoni.models",
"name": models.CharField(max_length=100),
"address": models.CharField(max_length=100),
}
In [26]: Person = type('Person', (models.Model,), attributes)
In [27]: Person
Out[27]: class 'lekhoni.models.Person'
In [28]: p1= Person()
In [29]: p1.name= 'manir'
In [30]: p1.save()
In [31]: Person.objects.a
Person.objects.aggregate Person.objects.all Person.objects.annotate
In [32]: Person.objects.all()
Out[33]: [Person: Person object]
It seems very simple- not sure why it should not be a considered an option- Reflection is very common is other languages like C# or Java- Anyway I am very new to django things-
Are you talking about in a front end interface, or in the Django admin?
You can't create real fields on the fly like that without a lot of work under the hood. Each model and field in Django has an associated table and column in the database. To add new fields usually requires either raw sql, or migrations using South.
From a front end interface, you could create pseudo fields, and store them in a json format in a single model field.
For example, create an other_data text field in the model. Then allow users to create fields, and store them like {'userfield':'userdata','mileage':54}
But I think if you're using a finite class like vehicles, you would create a base model with the basic vehicle characteristics, and then create models that inherits from the base model for each of the vehicle types.
class base_vehicle(models.Model):
color = models.CharField()
owner_name = models.CharField()
cost = models.DecimalField()
class car(base_vehicle):
mileage = models.IntegerField(default=0)
etc