Assume I have a model named MyModel and I have a Field Named field Now I want to add three more fields inside the prescription like one field_a , field_b and field_c .
Does Django Allow that in any way or we have to make another model for this purpose and then link with Foreign Key to MyModel?
Well I think it is an idea that could lead to some really hard to maintain code and should be well thought through.
That aside If you have a look at:
https://docs.wagtail.io/en/stable/topics/streamfield.html
It's a special field meant for cms applications that uses json in the field to model dynamic fields and validations for it.
Also if you run postgres this might help. https://docs.djangoproject.com/en/3.2/ref/contrib/postgres/fields/#jsonfield
Here is my models.py:
class Item(models.Model):
# ... some irrelevent fields ...
tags = models.ManyToManyField('Tag')
class Tag(models.Model):
name = models.CharField(max_lenght=30)
category_id = models.IntegerField()
Tag is actually a general-purpose name. Each item has many different type of tags - currently there are four types: team tags, subject tags, admin tags and special tags. eventually there will probably be a few more.
The idea is, they all have basically the same fields, so instead of having like 4 tables with manytomany relationship, and instead of adding a new column for Item whenever adding a new type, everything is called a 'tag' and it's very easy to add new types without any change to the schema.
Now to handle this in the admin.py I'm using dyamically created proxy models (based on this), as such:
def create_modeladmin(modeladmin, model, name = None):
class Meta:
proxy = True
app_label = model._meta.app_label
attrs = {'__module__': '', 'Meta': Meta}
newmodel = type(name, (model,), attrs)
admin.site.register(newmodel, modeladmin)
return modeladmin
class TagAdmin(models.Model):
def queryset(self):
return self.model.objects.filter(category_id = self.cid)
class TeamAdmin(TagAdmin):
cid = 1
class SubjectAdmin(TagAdmin):
cid = 2
# ... and so on ...
create_modeladmin(TeamAdmin, name='Teams', model=Tag)
create_modeladmin(SubjectAdmin, name='Subject', model=Tag)
#... and so on ...
This works great for me. However, different staff members need different editing permissions - one guy shouldn't access admin tags, while the other should only have access to edit subject-tags and team-tags. But as far as the admin site is concerned - the dynamic models do not exist in the permission list, and I can't give anyone permissions regarding them.
i.e. a user given all permissions on the list will still not have access to edit any of the dynamic models, and the only way to let anyone access them at all is to give him a superuser which obviously defies the point
I searched SO and the web and I can't anyone with a similar problem, and the docs don't say anything about this not in the dynamic models section or the proxy models section. so I'm guessing this is a different kind of problem. Any help would be greatly appreciated
UPDATE
So after some research into it, the answer was simple enough. Since permissions in django are objects that are saved to the database, what I needed to do was simple - add the relevent permissions (and create new ContentType objects as well) to the db, and then I could give people specific pemissions.
However, this raised a new question - is it a good convention to put the function that creates the permissions inside create_modeladmin as a find_or_create sort of function (that basically runs every time) or should it be used as an external script that I should run once every time I add a new dynamic model (sort of like how syncdb does it)?
And is there a way to also create the permissions dynamically (which seems to me like the ideal and most fitting solution)?
of course you can create permissions, django have django.contrib.auth.management.create_permissions to do this
So in my project I am trying to extend the User model to a Staff class and the Group model to a PermGroup class. However, when I save a PermGroup in the Staff's groups field (inherited from User) it only saves the PermGroup object as a Group and all of the fields and methods I defined in my PermGroup class are stripped away. So I decided the best course of action would be to override the groups field. From an earlier stackoverflow question I found and Django documentation, this should work.
class Staff(User):
User.groups = models.ManyToManyField('PermGroup', blank=True)
I need to use 'PermGroup' because the class shows up later in the file, and PermGroup has a field that relies on the Staff class, so if i switched the order I would have the same problem, only in the PermGroup class.
Now the problem I am having is that groups is now a ManyToManyField object where all the other "manytomany" fields are ManyRelatedManagers. I want groups to be a ManyRelatedManager but I do not know how.
Is it possible to get groups to be a ManyRelatedManager when I initiatize it using the 'PermGroup' model call?
If my approach is wrong and you can suggest an alternative to saving PermGroups in the Staff class. I would greatly appreciate it.
Why not just have your Staff be a standard model with a ForeignKey (OneToOneField, to be more exact) to his/her corresponding User?
And, to remove the circular dependency problem, you just need to make one dependent on the other. For instance, the PermGroup model could have a field of a ManytoMany of the Staff members in that group. There's no need for Staff to have a PermGroup, because if you wanted to see what groups a member belongs to, you'd just do something like this:
groups_theyre_in = PermGroups.objects.filter(staff_members__id=id_were_looking_for)
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
I have some newbie questions about Django.
I want to write a generic ticket-management system, where the administrator of the site should be able to add custom fields to a ticket. It seems that the database tables are generated on initialization, so it is not clear to me how to add custom fields at runtime.
One way is to have a long list of fields of different types, all nullable, and let the administrator rename/select the fields she needs. Is there a better design?
Thanks!
I'm currently in charge of maintaining a similar site where a treatment for a medical condition is listed and there can be arbitrary number of "cases" which are user-posted experiences for that treatment/condition combo attached.
The method my company used to set it up was to have an Entry object which would be analogous to the custom field you described, which has a Foreign Key referencing the treatment/condition to which it belongs.
Then when we want to get all the entries for a particular treatment/condition combo, we simply do an
Entry.objects.filter(condition=ID)
So, in your case, I would suggest having a Ticket model, and an "Entry" style model which contains a Foreign Key reference to the Ticket to which it belongs.
I would make something like the code below. Store extra attributes in an attribute model. Store the values in AttributeValue.
class Ticket(models.Model):
name = models.CharField(max_length=200)
class Attribute(models.Model):
name = models.CharField(max_length=200)
class AttributeValues(models.Model):
attribute = models.ForeignKey(Attribute)
ticket = models.ForeignKey(Ticket)
value = models.CharField(max_length=200)