Develop Reference

Stop affecting other objects of Django Many to Many model

I'm trying to replicate Blood Group as Model as defined in this picture.
.
In my models.py file I had my code to replicate the blood groups like this
class BloodGroup(models.Model):
name = models.CharField(
max_length=3
)
gives = models.ManyToManyField("self")
receives = models.ManyToManyField("self")
def __str__(self):
return self.name
And in my admin.py file I had registered the model as follows
class BloodGroupAdmin(admin.ModelAdmin):
model = BloodGroup
list_display = ['name', 'get_gives', 'get_receives']
def get_gives(self, obj):
return ", ".join([item.name for item in obj.gives.all()])
def get_receives(self, obj):
return ", ".join([item.name for item in obj.receives.all()])
admin.site.register(BloodGroup, BloodGroupAdmin)
Initially I created plain BloodGroup objects without their gives and receives attribute by providing just their names alone. Thus I create an object for all 8 types. Then as I added relationships to each object I found that adding gives or receives for one object affects other objects gives and receives too, making it impossible to replicate the structure in image.
How do I define relationships, without affecting other objects?
In my admin site, I see field names as "get_gives" and "get_receives". How would i make the admin page show field names as "gives" and "receives" but still displaying objects as strings like the image below?

For first question, probably it is better to have only one relation gives. receives can be found from the reverse query. Like this:
class BloodGroup(models.Model):
name = models.CharField(
max_length=3
)
gives = models.ManyToManyField("self", related_name="receives", symmetrical=False)
Then you only need to add objects to gives. receives will be generated automatically.
For second question, add short_description attribute to function(reference to docs). Like this:
get_gives.short_description = 'Gives'
get_receives.short_description = 'Receives'

How do I build a Django model that retrieves some fields from an API?

Question
How can I build a Model that that stores one field in the database, and then retrieves other fields from an API behind-the-scenes when necessary?
Details:
I'm trying to build a Model called Interviewer that stores an ID in the database, and then retrieves name from an external API. I want to avoid storing a copy of name in my app's database. I also want the fields to be retrieved in bulk rather than per model instance because these will be displayed in a paginated list.
My first attempt was to create a custom Model Manager called InterviewManager that overrides get_queryset() in order to set name on the results like so:
class InterviewerManager(models.Manager):
def get_queryset(self):
query_set = super().get_queryset()
for result in query_set:
result.name = 'Mary'
return query_set
class Interviewer(models.Model):
# ID provided by API, stored in database
id = models.IntegerField(primary_key=True, null=False)
# Fields provided by API, not in database
name = 'UNSET'
# Custom model manager
interviewers = InterviewerManager()
However, it seems like the hardcoded value of Mary is only present if the QuerySet is not chained with subsequent calls. I'm not sure why. For example, in the django shell:
>>> list(Interviewer.interviewers.all())[0].name
'Mary' # Good :)
>>> Interviewer.interviewers.all().filter(id=1).first().name
'UNSET' # Bad :(
My current workaround is to build a cache layer inside of InterviewManager that the model accesses like so:
class InterviewerManager(models.Manager):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
self.api_cache = {}
def get_queryset(self):
query_set = super().get_queryset()
for result in query_set:
# Mock querying a remote API
self.api_cache[result.id] = {
'name': 'Mary',
}
return query_set
class Interviewer(models.Model):
# ID provided by API, stored in database
id = models.IntegerField(primary_key=True, null=False)
# Custom model
interviewers = InterviewerManager()
# Fields provided by API, not in database
#property
def name(self):
return Interviewer.interviewers.api_cache[self.id]['name']
However this doesn't feel like idiomatic Django. Is there a better solution for this situation?
Thanks

why not just make the API call in the name property?
#property
def name(self):
name = get_name_from_api(self.id)
return name

If that isnt possible by manipulating a get request where you can add a list of names and recieve the data. The easy way is to do it is in a loop.
I would recommand you to build a so called proxy where you load the articles in a dataframe/dict, save this varible data ( with for example pickle ) and use it when nessary. It reduces loadtimes and is near efficient.

How to map an existing python class to a Django model

I'm writing a web scraper to get information about customers and appointment times to visit them. I have a class called Job that stores all the details about a specific job. (Some of its attributes are custom classes too e.g Client).
class Job:
def __init__(self, id_=None, client=Client(None), appointment=Appointment(address=Address(None)), folder=None,
notes=None, specific_reqs=None, system_notes=None):
self.id = id_
self.client = client
self.appointment = appointment
self.notes = notes
self.folder = folder
self.specific_reqs = specific_reqs
self.system_notes = system_notes
def set_appointment_date(self, time, time_format):
pass
def set_appointment_address(self, address, postcode):
pass
def __str__(self):
pass
My scraper works great as a stand alone app producing one instance of Job for each page of data scraped.
I now want to save these instances to a Django database.
I know I need to create a model to map the Job class onto but that's where I get lost.
From the Django docs (https://docs.djangoproject.com/en2.1/howto/custom-model-fields/) it says in order to use my Job class in the Django model I don't have to change it at all. That's great - just what I want. but I can't follow how to create a model that maps to my Job class.
Should it be something like
from django.db import models
import Job ,Client
class JobField(models.Field):
description = "Job details"
def __init__(self, *args, **kwargs):
kwargs['id_'] = Job.id_
kwargs['client'] = Client(name=name)
...
super().__init__(*args, **kwargs)
class Job(models.Model):
job = JobField()
And then I'd create a job using something like
Job.objects.create(id_=10101, name="Joe bloggs")
What I really want to know is am I on the right lines? Or (more likely) how wrong is this approach?
I know there must be a big chunk of something missing here but I can't work out what.

By mapping I'm assuming you want to automatically generate a Django model that can be migrated in the database, and theoretically that is possible if you know what field types you have, and from that code you don't really have that information.
What you need to do is to define a Django model like exemplified in https://docs.djangoproject.com/en/2.1/topics/db/models/.
Basically you have to create in a project app's models.py the following class:
from django import models
class Job(models.Model):
client = models.ForeignKey(to=SomeClientModel)
appointment = models.DateTimeField()
notes = models.CharField(max_length=250)
folder = models.CharField(max_length=250)
specific_reqs = models.CharField(max_length=250)
system_notes = models.CharField(max_length=250)
I don't know what data types you actually have there, you'll have to figure that out yourself and cross-reference it to https://docs.djangoproject.com/en/2.1/ref/models/fields/#model-field-types. This was just an example for you to understand how to define it.
After you have these figured out you can do the Job.objects.create(...yourdata).
You don't need to add an id field, because Django creates one by default for all models.

Django : Migration of polymorphic models back to a single base class

Let's suppose I have a polymorphic model and I want to get rid of it.
class AnswerBase(models.Model):
question = models.ForeignKey(Question, related_name="answers")
response = models.ForeignKey(Response, related_name="answers")
class AnswerText(AnswerBase):
body = models.TextField(blank=True, null=True)
class AnswerInteger(AnswerBase):
body = models.IntegerField(blank=True, null=True)
When I want to get all the answers I can never access "body" or I need to try to get the instance of a sub-class by trial and error.
# Query set of answerBase, no access to body
AnswerBase.objects.all()
question = Question.objects.get(pk=1)
# Query set of answerBase, no access to body (even with django-polymorphic)
question.answers.all()
I don't want to use django-polymorphic because of performances, because it does not seem to work for foreignKey relation, and because I don't want my model to be too complicated. So I want this polymorphic architecture to become this simplified one :
class Answer(models.Model):
question = models.ForeignKey(Question, related_name="answers")
response = models.ForeignKey(Response, related_name="answers")
body = models.TextField(blank=True, null=True)
The migrations cannot be created automatically, it would delete all older answers in the database. I've read the Schema Editor documentation but it does not seem there is a buildin to migrate a model to something that already exists. So I want to create my own operation to save the AnswerText and AnswerInteger as an Answer then delete AnswerText and AnswerInteger. I'm hoping I won't have to write SQL directly, but maybe that's the only solution ? My migration file looks like this. I created an Operation called MigrateAnswer :
from myapp.migrations import MigrateAnswer
class Migration(migrations.Migration):
operations = [
migrations.RenameModel("AnswerBase", "Answer"),
migrations.AddField(
model_name='answer',
name='body',
field=models.TextField(blank=True, null=True),
),
MigrateAnswer("AnswerInteger"),
MigrateAnswer("AnswerText"),
migrations.DeleteModel(name='AnswerInteger',),
migrations.DeleteModel(name='AnswerText',),
]
So what I want to do in MigrateAnswer is to migrate the value for an old model (AnswerInteger and AnswerText) to the base class (now named Answer, previousely AnswerBase). Here's my operation class :
from django.db.migrations.operations.base import Operation
class MigrateAnswer(Operation):
reversible = False
def __init__(self, model_name):
self.old_name = model_name
def database_forwards(self, app_label, schema_editor, from_state,
to_state):
new_model = to_state.apps.get_model(app_label, "Answer")
old_model = from_state.apps.get_model(app_label, self.old_name)
for field in old_model._meta.local_fields:
# loop on "question", "reponse" and "body"
# schema_editor.alter_field() Alter a field on a single model
# schema_editor.add_field() + remove_field() Does not permit
# to migrate the value from the old field to the new one
pass
So my question is : Is it possible to do this wihout using "execute" (ie : without writing SQL). If so what should I do in the for loop of my Operation ?
Thanks in advance !

There is no need to write an Operations class; data migrations can be done simply with a RunPython call, as the docs show.
Within that function you can use perfectly normal model instance methods; since you know the fields you want to move the data for, there is no need to get them via meta lookups.
However you will need to temporarily call the new body field a different name, so it doesn't conflict with the old fields on the subclasses; you can rename it back at the end and delete the child classes because the value will be in the base class.
def migrate_answers(apps, schema_editor):
classes = []
classes_str = ['AnswerText', 'AnswerInteger']
for class_name in classes_str:
classes.append(apps.get_model('survey', class_name))
for class_ in classes:
for answer in class_.objects.all():
answer.new_body = answer.body
answer.save()
operations = [
migrations.AddField(
model_name='answerbase',
name='new_body',
field=models.TextField(blank=True, null=True),
),
migrations.RunPython(migrate_answers),
migrations.DeleteModel(name='AnswerInteger',),
migrations.DeleteModel(name='AnswerText',),
migrations.RenameField('AnswerBase', 'new_body', 'body'),
migrations.RenameModel("AnswerBase", "Answer"),
]

You could create an empty migration for the app you want to do these modifications and use the migrations.RunPython Class to execute custom python functions.
Inside these functions you can have access to your models
The Django ORM that you can do data manipulation.
Pure python, no raw SQL.

how to modify related object's data whilie creating a new object in django

I am new to django and I'm trying to do something pretty simple.
my models.py is as below:
from django.db import models
class DiskDrive(models.Model):
deviceId = models.CharField(max_length=64, primary_key=True)
freeSpace = models.BigIntegerField()
def __unicode__(self):
return self.deviceId
class StoragePool(models.Model):
poolId = models.CharField(max_length=256, primary_key=True)
size = models.BigIntegerField()
drive = models.ForeignKey(DiskDrive, related_name='pools')
def __unicode__(self):
return self.poolId
I haven't added anything to views.py and urls.py yet.
I'm able to create objects of both the classes.
Whenever I create an object of StoragePool class, I want to reduce the value of 'freeSpace' attribute of the related DiskDrive object by the 'size' of 'StoragePool' object. How should I do this? Please help...

Sounds like the perfect job for a post_save signal:
#receiver(post_save, sender=StoragePool)
def update_drive_space(sender, instance, created, **kwargs):
if created:
instance.drive.freeSpace = F('freeSpace') - instance.size
instance.drive.save(update_fields=['freeSpace'])
See Django signals documentation for more info about how signals work. In a nutshell this method will be called each time you create or update a StoragePool object.
Few notes:
I am using F expression to reference current database size value instead of blindly saving whatever we have on Django side (this will ensure correct value when multiple clients will create new StoragePool objects)
It is likely that you want to adjust freeSpace attribute also when size is updated - not just on StoragePool creation. To make that happen just delete if created check and it will be run on every StoragePool.save()