SQLALchemy update ARRAY column [duplicate] - sqlalchemy

I'm working on a project using Flask and a PostgreSQL database, with SQLAlchemy.
I have Group objects which have a list of User IDs who are members of the group. For some reason, when I try to add an ID to a group, it will not save properly.
If I try members.append(user_id), it doesn't seem to work at all. However, if I try members += [user_id], the id will show up in the view listing all the groups, but if I restart the server, the added value(s) is (are) not there. The initial values, however, are.
Related code:
Adding group to the database initially:
db = SQLAlchemy(app)
# ...
g = Group(request.form['name'], user_id)
db.session.add(g)
db.session.commit()
The Group class:
from flask.ext.sqlalchemy import SQLAlchemy
from sqlalchemy.dialects.postgresql import ARRAY
class Group(db.Model):
__tablename__ = "groups"
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(128))
leader = db.Column(db.Integer)
# list of the members in the group based on user id
members = db.Column(ARRAY(db.Integer))
def __init__(self, name, leader):
self.name = name
self.leader = leader
self.members = [leader]
def __repr__(self):
return "Name: {}, Leader: {}, Members: {}".format(self.name, self.leader, self.members)
def add_user(self, user_id):
self.members += [user_id]
My test function for updating the Group:
def add_2_to_group():
g = Group.query.all()[0]
g.add_user(2)
db.session.commit()
return redirect(url_for('show_groups'))
Thanks for any help!

As you have mentioned, the ARRAY datatype in sqlalchemy is immutable. This means it isn’t possible to add new data into array once it has been initialised.
To solve this, create class MutableList.
from sqlalchemy.ext.mutable import Mutable
class MutableList(Mutable, list):
def append(self, value):
list.append(self, value)
self.changed()
#classmethod
def coerce(cls, key, value):
if not isinstance(value, MutableList):
if isinstance(value, list):
return MutableList(value)
return Mutable.coerce(key, value)
else:
return value
This snippet allows you to extend a list to add mutability to it. So, now you can use the class above to create a mutable array type like:
class Group(db.Model):
...
members = db.Column(MutableList.as_mutable(ARRAY(db.Integer)))
...

You can use the flag_modified function to mark the property as having changed. In this example, you could change your add_user method to:
from sqlalchemy.orm.attributes import flag_modified
# ~~~
def add_user(self, user_id):
self.members += [user_id]
flag_modified(self, 'members')

To anyone in the future: so it turns out that arrays through SQLAlchemy are immutable. So, once they're initialized in the database, they can't change size. There's probably a way to do this, but there are better ways to do what we're trying to do.

This is a hacky solution, but what you can do is:
Store the existing array temporarily
Set the column value to None
Set the column value to the existing temporary array
For example:
g = Group.query.all()[0]
temp_array = g.members
g.members = None
db.session.commit()
db.session.refresh(g)
g.members = temp_array
db.session.commit()

In my case it was solved by using the new reference for storing a object variable and assiging that new created variable in object variable.so, Instead of updating the existing objects variable it will create a new reference address which reflect the changes.
Here in Model,
Table: question
optional_id = sa.Column(sa.ARRAY(sa.Integer), nullable=True)
In views,
option_list=list(question.optional_id if question.optional_id else [])
if option_list:
question.optional_id.clear()
option_list.append(obj.id)
question.optional_id=option_list
else:
question.optional_id=[obj.id]

Related

django admin site returns MultipleObjectsReturned exception with inspectdb imported legacy database and composite primary key

Using inspectdb, I have imported a legacy database, that contains entities with composite primary keys, in django . The database schema contains about 200 different entities and inspectdb is quite handy in that situation.
This is the schema in mysql:
CREATE TABLE `mymodel` (
`id` bigint(20) unsigned NOT NULL DEFAULT '0',
`siteid` bigint(20) unsigned NOT NULL DEFAULT '0',
...
PRIMARY KEY (`siteid`,`id`),
...
Following the autogenerated model in django (imported using python manager.py inspectdb)
class Mymodel(models.Model):
id = models.PositiveBigIntegerField()
siteid = models.PositiveBigIntegerField(primary_key=True)
...
class Meta:
managed = False
db_table = 'mymodel'
unique_together = (('siteid', 'id'),
I have registered all models in the admin site using the following approach:
from django.contrib import admin
from django.apps import apps
app = apps.get_app_config('appname')
for model_name, model in app.models.items():
admin.site.register(model)
After all the work is done, I navigate to the admin site and click on any object in the "mymodel" section and the following exception will be returned:
appname.models.Content.MultipleObjectsReturned: get() returned more than one Mymodel-- it returned more than 20!
Obviously, (this is what it seems to me at least) admin is using the siteid to get the object, tough it should use the unique_together from the Meta class.
Any suggestions how I can achieve to solve this with a general configuration and get the admin site module to query using the unique_together?
Yes you can solve this problem but you put a little more effort.
First you separate model-admin class for model Mymodel and customize model-admin class method:
Since django admin build change url in ChangeList class, So we can create a custom Changelist class like MymodelChangelist and pass id field value as a query params. We will use id field value to getting object.
Override get_object() method to use custom query for getting object from queryset
Override get_changelist() method of model-admin to set your custom Changelist class
Override save_model() method to save object explicitly.
admin.py
class MymodelChangelist(ChangeList):
# override changelist class
def url_for_result(self, result):
id = getattr(result, 'id')
pk = getattr(result, self.pk_attname)
url = reverse('admin:%s_%s_change' % (self.opts.app_label,
self.opts.model_name),
args=(quote(pk),),
current_app=self.model_admin.admin_site.name)
# Added `id` as query params to filter queryset to get unique object
url = url + "?id=" + str(id)
return url
#admin.register(Mymodel)
class MymodelAdmin(admin.ModelAdmin):
list_display = [
'id', 'siteid', 'other_model_fields'
]
def get_changelist(self, request, **kwargs):
"""
Return the ChangeList class for use on the changelist page.
"""
return MymodelChangelist
def get_object(self, request, object_id, from_field=None):
"""
Return an instance matching the field and value provided, the primary
key is used if no field is provided. Return ``None`` if no match is
found or the object_id fails validation.
"""
queryset = self.get_queryset(request)
model = queryset.model
field = model._meta.pk if from_field is None else model._meta.get_field(from_field)
try:
object_id = field.to_python(object_id)
# get id field value from query params
id = request.GET.get('id')
return queryset.get(**{'id': id, 'siteid': object_id})
except (model.DoesNotExist, ValidationError, ValueError):
return None
def save_model(self, request, obj, form, change):
cleaned_data = form.cleaned_data
if change:
id = cleaned_data.get('id')
siteid = cleaned_data.get('siteid')
other_fields = cleaned_data.get('other_fields')
self.model.objects.filter(id=id, siteid=siteid).update(other_fields=other_fields)
else:
obj.save()
Now you can update any objects and also add new object. But, On addition one case you can't add- siteid which is already added because of primary key validation

Is there a way to set the id of an existing instance as the value of a nested serializer in DRF?

I'm developing a chat application. I have a serializer like this:
class PersonalChatRoomSerializer(serializers.ModelSerializer):
class Meta:
model = PersonalChatRoom
fields = '__all__'
user_1 = UserSerializer(read_only=True)
user_2 = UserSerializer()
the user_1 field is auto-populated but the client should provide the user_2 field in order to create a personal chat room with another user.
My problem is, when creating a new chat room, the serializer tries to create a new user object from the input data thus giving me validation errors. What I really want it to do is to accept a user id and set the value of user_2 field to an existing user instance that is currently available in the database and if the user is not found, simply return a validation error. (the exact behavior of PrimaryKeyRelatedField when creating a new object)
I want my input data to look like this:
{
'user_2': 1 // id of the user
}
And when I retrieve my PersonalChatRoom object, I want the serialized form of the user object for my user_2 field:
{
...,
'user_2': {
'username': ...,
'the_rest_of_the_fields': ...
}
}
How can I achieve this?
views.py
class GroupChatRoomViewSet(viewsets.ModelViewSet):
permission_classes = [IsUserVerified, IsGroupOrIsAdminOrReadOnly]
serializer_class = GroupChatRoomSerializer
def get_queryset(self):
return self.request.user.group_chat_rooms.all()
def perform_create(self, serializer):
return serializer.save(owner=self.request.user)
I finally figured out how to do it. I just needed to override the to_representation method and serialize the object there. Here is the code I ended up with:
class PersonalChatRoomSerializer(serializers.ModelSerializer):
class Meta:
model = PersonalChatRoom
fields = '__all__'
read_only_fields = ['user_1']
def to_representation(self, chat_room):
""" Serialize user instances when outputing the results """
obj = super().to_representation(chat_room)
for field in obj.keys():
if field.startswith('user_'):
obj[field] = UserSerializer(User.objects.get(pk=obj[field])).data
return obj

Persist one object from one database to another using sqlalchemy

I have two databases (both Mysql) that have exactly the same tables, and I want to copy some data from one to another using Sqlalchemy.
I can copy simple objects following the answer given in this question:
Cannot move object from one database to another
The problem is when the object has dependencies from another table, and I want to copy the dependencies as well.
So to make it more clear, this is my model (the same for both databases but using a different bind_key that points to a different database):
db1 = SQLAlchemy()
Class Payment(db.Model):
__tablename__ = 'payments'
__bind_key__ = 'db1'
id = db1.Column(db.Integer, primary_key=True)
paymethod_id = db1.Column(db.Integer(), db1.ForeignKey(PaymentMethod.id))
payment_method = db1.relationship(PaymentMethod)
What I would like to do is the following:
from models1 import Payment as Payment1
from models2 import Payment as Payment2
# query from one database
payment1 = db1.session.query(Payment1).first()
# create and add it to the other database
payment2 = Payment2(**payment1.__dict__.copy())
db2.session.add(payment)
db2.session.commit()
But in this case the foreign key fails because I don't have the PaymentMethod stored yet.
Is there a different approach to do that or I would have to do this procedure for every dependency of my object and be sure that I store the children beforehand?
Any help is appreciated :)
I came up with a solution that remaps the object to the right model and stores all its children. You call the method save_obj and pass the object you want to map. It will then retrieve a table with the same name but then from the model you want to remap the object to and it will recursively do the same for all its children. You have to define the right model in the method get_model.
To run this is necessary to disable autoflush to prevent committing before the object is correctly formed and it is also necessary to commit after calling the method. I'm using flask-sqlalchemy.
Hope this can help or give some insight to someone that faces a similar problem :)
def save_obj(obj, checked=[]):
if obj in checked:
# if the object was already converted, retrieve the right object
model = get_model(obj.__mapper__.mapped_table.name)
return get_obj(obj, model)
checked.append(obj)
children = []
relations = obj.__mapper__.relationships.items()
# get all the relationships of this model (foreign keys)
for relation in relations:
model = get_model(relation[1].table.name)
if model:
# remove the cascade option for this object, so the children are not stored automatically in the session
relation[1]._cascade = CascadeOptions('')
child = getattr(obj, relation[0])
if not child:
continue
# if the child is a set of children
if isinstance(child, list):
new_children = []
for ch in copy(child):
# convert the child
new_child = save_obj(ch, checked)
new_children.append(new_child)
children.append((relation[0], new_children))
else:
new_child = save_obj(child, checked)
children.append((relation[0], new_child))
# get the model of the object passed
model = get_model(obj.__mapper__.mapped_table.name)
new_obj = get_obj(obj, model)
# set all the children in this object
for child in children:
if child[1]:
setattr(new_obj, child[0], child[1])
checked.append(new_obj)
session.add(new_obj)
return new_obj
def get_model(table_name):
# get the right model for this object
for table in db.Model._decl_class_registry.values():
if hasattr(table, '__tablename__') and table.__tablename__ == table_name:
return table
return None
def create_new_obj(obj, model):
params = obj.__dict__.copy()
params.pop('_sa_instance_state')
return model(**params)
def get_obj(child, model):
# check if the object is already stored in the db
child_in_db = session.query(model).get(child.id)
if child_in_db:
return child_in_db
# check if the object is already in the session
for s in session.new:
if type(s) == model and s.id == child.id:
return s
return create_new_obj(child, model)

How to build backref with both associatition object and secondaryjoin?

I need some models for instance following:
Work - e.g. works of literature.
Worker - e.g. composer, translator or something similar has contribution to work.
Thus, a 'type' field is required to distinguish workers by division of work. As SQLAlchemy's documentation, this case can benifit from association object like following:
class Work(base):
id = Column(Integer, primary_key=True)
name = Column(String(50))
description = Column(Text)
class Worker(base):
id = Column(Integer, primary_key=True)
name = Column(String(50))
description = Column(Text)
class Assignment(base):
work_id = Column(Integer, Foreignkey('work.id'), primary_key=True)
worker_id = Column(Integer, Foreignkey('worker.id'), primary_key=True)
type = Column(SmallInteger, nullable=True)
Nonetheless, how to take advantage of backref and alternatvie join condition for building relation immediately to implement that each Work object can retrieve and modify corresponding Worker(s) via different attributions for distinction. For example:
work = session.query(Work).get(1)
work.name
>>> 'A Dream of The Red Mansions'
work.composers
>>> [<Worker('Xueqin Cao')>]
work.translators
>>> [<Worker('Xianyi Yang')>, <Worker('Naidie Dai')>]
Vice versa:
worker = session.query(Worker).get(1)
worker.name
>>> 'Xueqin Cao'
worker.composed
>>> [<Work('A Dream of The Red Mansions')>]
worker.translated
>>> []
Adding secondaryjoin directly without secondary specified seems not feasible, besides, SQLAlchemy's docs notes that:
When using the association object pattern, it is advisable that the association-mapped table not be used as the secondary argument on a relationship() elsewhere, unless that relationship() contains the option viewonly=True. SQLAlchemy otherwise may attempt to emit redundant INSERT and DELETE statements on the same table, if similar state is detected on the related attribute as well as the associated object.
Then, is there some way to build these relations elegantly and readily ?
There's three general ways to go here.
One is, do a "vanilla" setup where you have "work"/"workers" set up without distinguishing on "type" - then, use relationship() for "composer", "composed", "translator", "translated" by using "secondary" to Assignment.__table__ along with custom join conditions, as well as viewonly=True. So you'd do writes via the vanilla properties only. A disadvantage here is that there's no immediate synchronization between the "vanilla" and "specific" collections.
Another is, same with the "vanilla" setup, but just use plain Python descriptors to give "composer", "composed", "translator", "translated" views in memory, that is, [obj.worker for obj in self.workers if obj.type == 'composer']. This is the simplest way to go. Whatever you put in the "vanilla" collections shows right up in the "filtered" collection, the SQL is simple, and there's fewer SELECT statements in play (one per Worker/Work instead of N per Worker/Work).
Finally, the approach that's closest to what you're asking, with primary joins and backrefs, but note with the association object, the backrefs are between Work/Assignment and Assignment/Worker, but not between Work/Worker directly. This approach probably winds up using more SQL to get at the results but is the most complete, and also has the nifty feature that the "type" is written automatically. We're also using a "one way backref", as Assignment doesn't have a simple way of relating back outwards (there's ways to do it but it would be tedious). Using a Python function to automate creation of the relationships reduces the boilerplate, and note here I'm using a string for "type", this can be an integer if you add more arguments to the system:
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy
Base = declarative_base()
def _work_assignment(name):
assign_ = relationship("Assignment",
primaryjoin="and_(Assignment.work_id==Work.id, "
"Assignment.type=='%s')" % name,
back_populates="work", cascade="all, delete-orphan")
assoc = association_proxy("%s_assign" % name, "worker",
creator=lambda worker: Assignment(worker=worker, type=name))
return assoc, assign_
def _worker_assignment(name):
assign_ = relationship("Assignment",
primaryjoin="and_(Assignment.worker_id==Worker.id, "
"Assignment.type=='%s')" % name,
back_populates="worker", cascade="all, delete-orphan")
assoc = association_proxy("%s_assign" % name, "work",
creator=lambda work: Assignment(work=work, type=name))
return assoc, assign_
class Work(Base):
__tablename__ = 'work'
id = Column(Integer, primary_key=True)
name = Column(String(50))
description = Column(Text)
composers, composer_assign = _work_assignment("composer")
translators, translator_assign = _work_assignment("translator")
class Worker(Base):
__tablename__ = 'worker'
id = Column(Integer, primary_key=True)
name = Column(String(50))
description = Column(Text)
composed, composer_assign = _worker_assignment("composer")
translated, translator_assign = _worker_assignment("translator")
class Assignment(Base):
__tablename__ = 'assignment'
work_id = Column(Integer, ForeignKey('work.id'), primary_key=True)
worker_id = Column(Integer, ForeignKey('worker.id'), primary_key=True)
type = Column(String, nullable=False)
worker = relationship("Worker")
work = relationship("Work")
e = create_engine("sqlite://", echo=True)
Base.metadata.create_all(e)
session = Session(e)
ww1, ww2, ww3 = Worker(name='Xueqin Cao'), Worker(name='Xianyi Yang'), Worker(name='Naidie Dai')
w1 = Work(name='A Dream of The Red Mansions')
w1.composers.append(ww1)
w1.translators.extend([ww2, ww3])
session.add(w1)
session.commit()
work = session.query(Work).get(1)
assert work.name == 'A Dream of The Red Mansions'
assert work.composers == [ww1]
assert work.translators == [ww2, ww3]
worker = session.query(Worker).get(ww1.id)
assert worker.name == 'Xueqin Cao'
assert worker.composed == [work]
assert worker.translated == []
worker.composed[:] = []
# either do this...
session.expire(work, ['composer_assign'])
# or this....basically need composer_assign to reload
# session.commit()
assert work.composers == []

Unique validator in WTForms with SQLAlchemy models

I defined some WTForms forms in an application that uses SQLALchemy to manage database operations.
For example, a form for managing Categories:
class CategoryForm(Form):
name = TextField(u'name', [validators.Required()])
And here's the corresponding SQLAlchemy model:
class Category(Base):
__tablename__= 'category'
id = Column(Integer, primary_key=True)
name = Column(Unicode(255))
def __repr__(self):
return '<Category %i>'% self.id
def __unicode__(self):
return self.name
I would like to add a unique constraint on the form validation (not on the model itself).
Reading the WTForms documentation, I found a way to do it with a simple class:
class Unique(object):
""" validator that checks field uniqueness """
def __init__(self, model, field, message=None):
self.model = model
self.field = field
if not message:
message = u'this element already exists'
self.message = message
def __call__(self, form, field):
check = self.model.query.filter(self.field == field.data).first()
if check:
raise ValidationError(self.message)
Now I can add that validator to the CategoryForm like this:
name = TextField(u'name', [validators.Required(), Unique(Category, Category.name)])
This check works great when the user tries to add a category that already exists \o/
BUT it won't work when the user tries to update an existing category (without changing the name attribute).
When you want to update an existing category : you'll instantiate the form with the category attribute to edit:
def category_update(category_id):
""" update the given category """
category = Category.query.get(category_id)
form = CategoryForm(request.form, category)
The main problem is I don't know how to access the existing category object in the validator which would let me exclude the edited object from the query.
Is there a way to do it? Thanks.
In the validation phase, you will have access to all the fields. So the trick here is to pass in the primary key into your edit form, e.g.
class CategoryEditForm(CategoryForm):
id = IntegerField(widget=HiddenInput())
Then, in the Unique validator, change the if-condition to:
check = self.model.query.filter(self.field == field.data).first()
if 'id' in form:
id = form.id.data
else:
id = None
if check and (id is None or id != check.id):
Although this is not a direct answer I am adding it because this question is flirting with being an XY Problem. WTForms primary job is to validate that the content of a form submission. While a decent case could be made that verifying that a field's uniqueness could be considered the responsibility of the form validator, a better case could be made that this is the responsibility of the storage engine.
In cases where I have be presented with this problem I have treated uniqueness as an optimistic case, allowed it to pass form submission and fail on a database constraint. I then catch the failure and add the error to the form.
The advantages are several. First it greatly simplifies your WTForms code because you do not have to write complex validation schemes. Secondly, it could improve your application's performance. This is because you do not have to dispatch a SELECT before you attempt to INSERT effectively doubling your database traffic.
The unique validator needs to use the new and the old data to compare first before checking if the data is unique.
class Unique(object):
...
def __call__(self, form, field):
if field.object_data == field.data:
return
check = DBSession.query(model).filter(field == data).first()
if check:
raise ValidationError(self.message)
Additionally, you may want to squash nulls too. Depending on if your truly unique or unique but allow nulls.
I use WTForms 1.0.5 and SQLAlchemy 0.9.1.
Declaration
from wtforms.validators import ValidationError
class Unique(object):
def __init__(self, model=None, pk="id", get_session=None, message=None,ignoreif=None):
self.pk = pk
self.model = model
self.message = message
self.get_session = get_session
self.ignoreif = ignoreif
if not self.ignoreif:
self.ignoreif = lambda field: not field.data
#property
def query(self):
self._check_for_session(self.model)
if self.get_session:
return self.get_session().query(self.model)
elif hasattr(self.model, 'query'):
return getattr(self.model, 'query')
else:
raise Exception(
'Validator requires either get_session or Flask-SQLAlchemy'
' styled query parameter'
)
def _check_for_session(self, model):
if not hasattr(model, 'query') and not self.get_session:
raise Exception('Could not obtain SQLAlchemy session.')
def __call__(self, form, field):
if self.ignoreif(field):
return True
query = self.query
query = query.filter(getattr(self.model,field.id)== form[field.id].data)
if form[self.pk].data:
query = query.filter(getattr(self.model,self.pk)!=form[self.pk].data)
obj = query.first()
if obj:
if self.message is None:
self.message = field.gettext(u'Already exists.')
raise ValidationError(self.message)
To use it
class ProductForm(Form):
id = HiddenField()
code = TextField("Code",validators=[DataRequired()],render_kw={"required": "required"})
name = TextField("Name",validators=[DataRequired()],render_kw={"required": "required"})
barcode = TextField("Barcode",
validators=[Unique(model= Product, get_session=lambda : db)],
render_kw={})
Looks like what you are looking for can easily be achieved with ModelForm which is built to handle forms that are strongly coupled with models (the category model in your case).
To use it:
...
from wtforms_components import Unique
from wtforms_alchemy import ModelForm
class CategoryForm(ModelForm):
name = TextField(u'name', [validators.Required(), Unique(Category, Category.name)])
It will verify unique values while considering the current value in the model. You can use the original Unique validator with it.
This worked for me, simple and easy:
Make sure that every time when a new row created in DB it must have unique name in colomn_name_in_db otherwise it will not work.
class SomeForm(FlaskForm):
id = IntegerField(widget=HiddenInput())
fieldname = StringField('Field name', validators=[DataRequired()])
...
def validate_fieldname(self, fieldname):
names_in_db = dict(Model.query.with_entities(Model.id,
Model.colomn_name_in_db).filter_by(some_filtes_if_needed).all())
if fieldname.data in names_in_db.values() and names_in_db[int(self.id)] != fieldname.data:
raise ValidationError('Name must be unique')