Develop Reference

Flask-Restless dump_to of primary key field

I am running into an issue that may be bug, but want to verify it with the community. I am basically trying to conform to camelcase for transporting data, then underscore for the database.
However, on the person_serializer, flask-restless will not allow an outbound "idPerson" as a result of the dump_to="idPerson". For some reason, it checks that the primary key exists and gets a keyError since the actual key is "id_person", not "idPerson".
Any help would be appreciated.
class Person(Base):
__tablename__ = "person"
id_person = Column(Integer, primary_key=True)
first_name = Column(String(50))
last_name = Column(String(50))
class PersonSchema(Schema):
id_person = fields.Integer(load_from="idPerson",dump_to="idPerson")
first_name = fields.String(load_from="firstName", dump_to="firstName")
last_name = fields.String(load_from="lastName", dump_to="lastName")
#post_load
def make_user(self, data):
return Person(**data)
person_schema = PersonSchema()
def person_serializer(instance):
return person_schema.dump(instance).data
def person_deserializer(data):
return person_schema.load(data).data
KEY ERROR IS BELOW
try:
# Convert the dictionary representation into an instance of the
# model.
instance = self.deserialize(data)
# Add the created model to the session.
self.session.add(instance)
self.session.commit()
# Get the dictionary representation of the new instance as it
# appears in the database.
result = self.serialize(instance)
except self.validation_exceptions as exception:
return self._handle_validation_exception(exception)
# Determine the value of the primary key for this instance and
# encode URL-encode it (in case it is a Unicode string).
pk_name = self.primary_key or primary_key_name(instance)
> primary_key = result[pk_name]
E KeyError: 'idPerson'

how to store only one value on my django models?

below is my model
from django.db import models
LEVEL_CHOICES = (('beg','beginner'),('ind','inter'),('exp','expert'))
class scrap(models.Model):
subject = models.CharField(max_length=128,unique=True)
# level= models.CharField(max_length=128,unique=True)
level = models.CharField(max_length=128, choices=LEVEL_CHOICES)
time = models.IntegerField(unique=True)
def __str__(self):
return self.subject
is there anyway so that i can only store one value per class.what i am doing is supplying values to this model's objects through a form.so i want only once instance of this class.if another value is supplied through through the form,what i want is the old value to be replaced and new value to be stored.

I'm not sure why do you really just keep one object for your model, but it's pretty easy to achieve:
if scrap.objects.exists():
current_scrap = Scrap.objects.all()[0]
else:
current_scrap = Scrap()
current_scrap.subject = new_subject
current_scrap.level = new_level
current_scrap.time = new_time
current_scrap.save()

django - can't assign a foreign key

For unknown reasons, I cannot assign a foreign key instance of Item_rarity table into Detailed_item table. Django throws an error:
Cannot assign "u'Basic'": "Detailed_item.rarity" must be a "Item_rarity" instance.
... But in Item_rarity dictionary "Basic" record exists - I can choose it from admin panel and create Detailed_item record manually.
I have defined models:
class Detailed_item(models.Model):
item_id = models.IntegerField(primary_key=True)
name = models.CharField(max_length=50)
level = models.IntegerField()
icon = models.CharField(max_length=150)
rarity = models.ForeignKey('Item_rarity')
general_type = models.ForeignKey('Item_type')
detailed_type = models.ForeignKey('Item_detailed_type')
class Item_rarity(models.Model):
name = models.CharField(max_length=15, primary_key=True)
class Item_type(models.Model):
name = models.CharField(max_length=15, primary_key=True)
class Item_detailed_type(models.Model):
name = models.CharField(max_length=20, primary_key=True)
In views, I try to populate it in this manner (inserting multiple items):
...
items = get_all_items() #get dict of items
for element in items:
tmp_det_type = ''
for key, val in element.iteritems():
#get 'detailed type' from inner dict
if key == "type":
tmp_det_type = val
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity=element['rarity'], #error
general_type=element['type'],
detailed_type=tmp_det_type,
)
item.save()
...
I even tried to hard code "Basic" string, but it doesn't work either.
* Solved *
Next two entries, that is Item_type and Item_detailed_type were also invalid.
Correct code:
from app.models import Detailed_item, Item_rarity, Item_type, Item_detailed_type
...
items = get_all_items() #get dict of items
for element in items:
tmp_det_type = ''
for key, val in element.iteritems():
#get 'detailed type' from inner dict
if key == "type":
tmp_det_type = val
#create objects with string values
obj_rarity = Item_rarity(name=element['rarity'])
obj_item_type = Item_type(name=element['type'])
obj_item_detailed_type = Item_detailed_type(name=tmp_det_type)
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity=obj_rarity,
general_type=obj_item_type,
detailed_type=obj_item_detailed_type,
)
item.save()
...

Item_rarity instance should be passed while storing Detailed_item object since Item_rarity is a foreign key related object in Detailed_item.
Its that you might have passed the Basic string instead of the <Basic Object> itself.
While creating an object in django using its ORM, any foreign_key related object should be provided with the instance itself instead of the id(pk) of the object, where as while fetching the data from the database you can use either of instance or the id(pk) of the instance.
class ParentModel(models.Model):
model_field = models.CharField(max_length=16)
class MyModel(models.Model):
some_field = models.ForeignKey('ParentModel')
parent_model = ParentModel.objects.create(model_field='some_data')
my_model = MyModel.objects.create(some_field=parent_model)
^^^^^^^^^^^^
Note here that the parent_model object itself is passed instead of the id
While fetching the data back,
parent_model = ParentModel.objects.get(model_field='some_data')
my_model = MyModel.objects.get(some_field=parent_model)
or
my_model = MyModel.objects.get(some_field=parent_model.id)
Both would work in case of data fetch.

You do not have to provide the related object on creation if you change the kwarg in to rarity_name:
item = Detailed_item(
item_id=element['id'],
name=element['name'],
level=element['level'],
icon=element['icon'],
rarity_name=element['rarity'], # no error
general_type=element['type'],
detailed_type=tmp_det_type,
)
I have only tested this with the regular id field (the auto pk) but it
should work with your primary key just fine.
E.g.
class SimpleModel(Model):
value = TextField(blank=True)
class ComplexModel(Model):
simple = ForeingKey(SimpleModel)
title = TextField(unique=True)
ComplexModel.objects.create(title='test', simple_id=1)

I need to query for a set of objects whose primary keys are contained inside of a list

As the title says, I need a way to perform this query. I have tried the following:
user_list_ids = []
user_lists = []
user_entries = OwnerEntry.objects.filter(name=request.user)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
However, I get an error on the last line: int() argument must be a string or a number, not 'ListEntry'
Here are the relevant models:
class OwnerEntry(models.Model):
name = models.CharField(max_length=32)
list_id = models.ForeignKey(ListEntry)
class Meta:
ordering = ('name',)
class ListEntry(models.Model):
name = models.CharField(max_length=64)
# active_date = models.DateTimeField('date of last list activity')
expire_date = models.DateField('date of expiration')
create_date = models.DateField('date created')

to answer your question directly, please note that you have a list_id rather than list as a ForeignKey name (OwnerEntry model). In order to actually extract the fk value, you should use list_id_id instead (or rename list_id to list ;))
Please also note that django supports object references, like so:
someowner = OwnerEntry.objects.get( ... )
ownerslist = someowner.listentry_set.all()
cheers!

You can define OwnerEntry's foreign key to ListEntry as :
list_id = models.ForeignKey(ListEntry, related_query_name='owner_entry')
and then do this one-liner in your code:
user_lists = ListEntry.objects.filter(owner_entry__name=request.user)
What this does is exactly filter every ListEntry which has at least one owner_entry whose name is equal to request.user's.
The redefinition of the foreign key is just for the sake of giving a nice name to the query attribute.
For more details on queries that work with backward relationships: https://docs.djangoproject.com/en/dev/topics/db/queries/#lookups-that-span-relationships

(Django) Cannot assign "u'1'": "StaffProfile.user" must be a "User" instance

I have a model like below:
class StaffProfile(models.Model):
user = models.ForeignKey(User)
maas = models.FloatField()
maas_gunu = models.CharField(max_length=5)
When I try to insert data with a code like below:
staffprofilesay = StaffProfile.objects.filter(user = user_id).count()
if staffprofilesay > 0:
staffprofile = StaffProfile.objects.get(user = user_id)
else:
staffprofile = StaffProfile()
staffprofile.user = user_id
staffprofile.maas = calisan_formu.cleaned_data["maas"]
staffprofile.maas_gunu = calisan_formu.cleaned_data["maas_gunu"]
staffprofile.save()
I get an error like this:
Cannot assign "u'1'": "StaffProfile.user" must be a "User" instance.
What am I supposed to do?
PS: I'm using Django's User model

You need to assign a User object e.g.
from django.contrib.auth.models import User
user = User.objects.get(id=user_id)
staffprofile.user = user

user needs to be an instance of the User model, not a unicode object (which is what you are passing it).

Yes you have to pass User instance in staffprofile.user = user_id user id place.
As #david-s pointed out in a comment, if you don't have a user instance, you have to fetch from DB with an additional query.
Instead you can directly do is
staffprofile.user_id = user_id because Django behind the scene append _id in table for foreign keys so staffprofile.user will end staffprofile.user_id