I'm using Piston to create an API for an application in Django.
I'll try to explain my doubt on an easy way.
Let's think I've got two models:
class Device(models.Model):
id = models.TextField(...)
class Person(models.Model):
name = models.TextField(...)
device = models.ForeigKey(Device)
Now, if I receive an url like this:
(r'^api/(?P<person_name>\w+)/(?P<device_id>\w+)$',handler),
I want to add a person to the DB and, to do that, I need to add a new Device to the DB, but, since handlers in Piston are linked to a Model, how can I add a device to the DB in the same handler?
I tried something like this:
class PersonHandler(BaseHandler):
allowed_methods= ('PUT')
model = Person
def create(self, request, person_name, device_id):
Device.objects.create(id=device_id)
d = Device.objets.get(id=device_id)
Person.objects.create(name=person_name,device=d)
return rc.CREATED
But I guess it won't work.
How can I do what I want to do?
Related
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.
I'm trying to create an API endpoint on my Django project to retrieve data from my frontend.
I'm using two DBs on my django project, the first one is a SQLite DB, the second one is a MongoDB database, the data i need to retrieve is on MongoDB.
Here is my model:
class tst(models.Model):
_id = models.CharField(max_length=100)
ticker = models.FloatField()
def save(self): # ALL the signature
super(Trade, self).save(using='dbtwo')
Here is my view:
class tstList(generics.ListCreateAPIView):
queryset = tst.objects.all()
serializer_class = tstSerializer
And the url:
path('tst/', views.tstList.as_view()),
Everything is alright here but when i try to open the API from my browser, i keep getting the following error:
OperationalError at /tst/
no such table: main_tst
I think this happens because it tries to look for the table tst on the first SQLite database, instead of looking for it on the MongoDB one. Is there any way to solve this? I thought that adding using='dbtwo' would do it, but it's not the right solution.
Every advice is appreciated!
You need to define the database that you are using in the queryset for your API view
class tstList(generics.ListCreateAPIView):
queryset = tst.objects.using('dbtwo').all()
serializer_class = tstSerializer
Even better than this, if the model will only ever use the other database, you can set up a router so that you do not have to set "using" every time
class MyRouter:
def db_for_read(model, **hints):
if model == tst:
return 'dbtwo'
def db_for_write(model, **hints):
if model == tst:
return 'dbtwo'
# In your settings
DATABASE_ROUTERS = ['path.to.MyRouter']
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.
Let's assume that I have following models:
class ScoutBook(models.Model):
troop = models.ForeignKey('Dictionary', limit_choices_to={'type' : 'Troop'}, related_name='+', blank=True, null=True)
class Dictionary(models.Model):
name = models.CharField(max_length=CHAR_FIELD_MAX_LEN, verbose_name="Nazwa")
active = models.BooleanField(verbose_name="Aktywny")
type = models.CharField(max_length=CHAR_FIELD_MAX_LEN, choices=DICTIONARY_CHOICES)
and I want to implement following logic:
when creating ScoutBook allow users to select only active troops, and when editing allow to select active troops or allow user to leave value unchanged (even if the troop is inactive). If I use limit_choices_to = {..., 'active' = True} troop that is inactive is absent from combo box in django admin.
So to be clear: let's assume that there are four troops in this system: Troop1, Troop2 and InactiveTroop, InactiveTroop2. On model creation I would like user to be able to choose Troop1 and Troop2. If model has troop field set to InactiveTroop2, I would like user to be able to choose between InactiveTroop2, Troop1 and Troop2.
I was looking at the django forms api and I didn't found obvious way do this. Moreover, in the application I'm developing there will be many such fields and many such models --- so solution must be pain free. I would rather not create new Form class for every model. I will be using mostly django admin to enable editing the database, and some read only views that will just list entries.
Ideally I would like to encapsulate this functionality in some custom field --- but fields have access to model instance on validate and save phase --- so I dont have access to it when I produce formfield.
This sounds like something you want to do in a form, not in the object itself. Create a ModelForm and override the ModelChoiceField like this:
from django import forms
class ScoutBookForm(forms.ModelForm):
troop = forms.ModelChoiceField(queryset=Troop.objects.filter(active=True))
class Meta:
model = ScoutBook
You can also override the clean method of ScoutBook to ensure it cannot ever be saved with an inactive Troop, though that may have some unintended consequences (e.g., you wouldn't be able to update a ScoutBook in the admin if the troop had gone inactive at some point in the past).
Well I had to hook into ModelForm creation. Attached Form inspects it's fields and if specific conditions are met it replaces model field queryset.
class DictionayModelForm(ModelForm):
def __init__(self, *largs, **kwargs):
super(DictionayModelForm, self).__init__(*largs, **kwargs)
if self.instance and self.instance.pk is not None:
for f in self.instance._meta.fields:
if isinstance(f, models.ForeignKey) and issubclass(f.rel.to, Dictionary):
model_field = self.fields[f.name]
value = getattr(self.instance, f.name, None)
if value and value not in model_field.choices:
model_field.queryset = Dictionary.objects.filter(Q(**f.rel.limit_choices_to) | Q(id = value.id))
I'm thinking about this for a while now,
I'm creating a chat application, in chat.models a class Room is specified, however, a Room can be related to anything in my project, since it uses a generic relation in it's foreign key.
Is there a way to know which model that Room is related knowing only the models name?
Like:
ctype = 'user'
related_to_user = Room.objects.filter(content_type=ctype)
The problem I'm having is, the code below is in a view:
doc = get_object_or_404(Document, id=id)
# get *or create* a chat room attached to this document
room = Room.objects.get_or_create(doc)
If I don't want to use Document model, if I want a model associated to a string, a string that can be anything, without having to write tons of if's to get a specific Model for the specific string. Is there a way to find a model just by it's 'name'?
Thanks
http://docs.djangoproject.com/en/dev/ref/contrib/contenttypes/#methods-on-contenttype-instances
user_type = ContentType.objects.get(app_label="auth", model="user")
user_type = ContentType.objects.get(model="user")
# but this can throw an error if you have 2 models with the same name.
Very similar to django's get_model
from django.db.models import get_model
user_model = get_model('auth', 'user')
To use your example exactly:
ctype = ContentType.objects.get(model='user')
related_to_user = Room.objects.filter(content_type=ctype)