I have three tables which are defined as:
class User(Base):
__tablename__ = 'users'
id = Column(Integer(10), primary_key=True)
firstname = Column(String(64))
surname = Column(String(64))
class SWMS(Base):
__tablename__ = 'swms'
id = Column(Integer(10), primary_key=True)
owner_id = Column(Integer(10), ForeignKey('users.id', ondelete='CASCADE'))
filename = Column(String(255))
swmowner = relationship('User', backref=backref('users'))
class SWM_perms(Base):
__tablename__ = 'swm_perms'
id = Column(Integer(10), primary_key=True)
swm_id = Column(Integer(10), ForeignKey('swms.id', ondelete='CASCADE'))
user_id = Column(Integer(10), ForeignKey('users.id', ondelete='CASCADE'))
swm = relationship('SWMS', backref=backref('swms'))
swmuser = relationship('User', backref=backref('swmusers'))
Essentially, the SWMS table is a table of document info where the owner_id defines the user who created the document. SWM_perms is a table that has a mapping of document id to user id - to define which users are allowed to see the document.
To produce a table of all documents which are either 1) owned by the user or 2) are viewable by the user, in SQL I would do:
select owner_id, users.firstname, users.surname, filename
from swms, swm_perms, users
where users.id=swms.owner_id and
((swms.id=swm_perms.swm_id and swm_perms.user_id = 27) or (owner_id = 27));
How would you define this query in sqlalchemy? I am familiar with the or_() function but the variants I am trying do not generate the correct objects.
cond1 = and_(SWMS.id==SWM_perms.swm_id,SWM_perms.user_id==27)
swms = DBSession.query(User,SWMS).filter(or_(cond1,SWMS.owner_id==27)).\
filter(User.id==SWMS.owner_id).all()
and then you can do a list comprehension to pull the fields you want:
details = [(u.firstname, s.filename, s.blob_key, s.last_modified) for u,s in swms]
Also worth noting you can use the '&' operator, in place of 'and_', in the body of the query. See the example (second code block) they give here:
http://docs.sqlalchemy.org/en/rel_1_0/core/sqlelement.html#sqlalchemy.sql.expression.and_
Related
I am trying to create multi (i.e. 2-column index) for a model based table. But I'd like not to give specific name for this index. I'd like that naming_convention and alembic revision --autogenerate would do it's job with naming index. So far I have code like this:
from sqlalchemy import MetaData
from sqlalchemy.ext.declarative import as_declarative
from sqlalchemy.schema import Index
metadata = MetaData(
naming_convention={
'pk': '%(table_name)s_pk',
'ix': '%(table_name)s_%(column_0_N_name)s_ix',
},
)
#as_declarative(metadata=metadata)
class Base:
pass
class Foo(Base):
id = Column(Integer, primary_key=True)
col1 = Column('Col1', Integer)
col2 = Column('Col2', DateTime)
Index(
metadata.naming_convention['ix'] % {
'table_name': Foo.__tablename__,
'column_0_N_name': Foo.col1.expression.name + "_" + Foo.col2.expression.name
},
Foo.col1,
Foo.col2,
)
So I'd like to avoid the 'creating name' part of code:
metadata.naming_convention['ix'] % {
'table_name': Foo.__tablename__,
'column_0_N_name': Foo.col1.expression.name + "_" + Foo.col2.expression.name
}
after more search there is very simple solution. According to github comment in SQLAlchemy issue if you would like to create index by Index() you silmply need to pass name=None argument, and fill arguments for columns.
so the code above should look like (part that stays the same):
from sqlalchemy import MetaData
from sqlalchemy.ext.declarative import as_declarative
from sqlalchemy.schema import Index
metadata = MetaData(
naming_convention={
'pk': '%(table_name)s_pk',
'ix': '%(table_name)s_%(column_0_N_name)s_ix',
},
)
#as_declarative(metadata=metadata)
class Base:
pass
Option 1 (declare multi column index out of table model)
class Foo(Base):
id = Column(Integer, primary_key=True)
col1 = Column('Col1', Integer)
col2 = Column('Col2', DateTime)
Index(None, Foo.col1, Foo.col2)
Option 2 (declare multi column index inside of table model)
class Foo(Base):
id = Column(Integer, primary_key=True)
col1 = Column('Col1', Integer)
col2 = Column('Col2', DateTime)
__table_args__ = (
Index(None, 'Col1', 'Col2'),
)
Then index name (in both options) would be:
Foo_Col1_Col2_ix
I have not found this kind of solution in SQLALchemy documentation (maybe there is?), but it's good that there are some answers on github issues in SQLAlchemy github :)
I have tried to think about how the following SQL query would be structured as a Django ORM query but I have had no luck in my multiple attempts. Can anyone help?
SELECT targets_genetarget.gene, count(targets_targetprediction.gene) as total
FROM targets_genetarget
LEFT OUTER JOIN targets_targetprediction on targets_targetprediction.gene =
targets_genetarget.gene
WHERE list_name LIKE %s
GROUP BY targets_genetarget.gene
class GeneTarget(models.Model):
list_name = models.CharField(max_length=100)
gene = models.CharField(max_length=50)
date_added = models.DateField(auto_now=True)
class Meta:
unique_together = (('list_name', 'gene'),)
def __str__(self):
return self.list_name
class TargetPrediction(models.Model):
specimen_id = models.CharField(max_length=100)
patient_peptide = models.ForeignKey(Peptide, on_delete=models.CASCADE, verbose_name="Peptide", related_name="predictions")
allele = models.ForeignKey(Allele, on_delete=models.CASCADE, verbose_name="Allele", related_name="predictions")
gene = models.CharField(max_length=50)
class Meta:
unique_together = (('specimen_id', 'patient_peptide', 'allele', 'gene'),)
def get_absolute_url(self):
return f'/samples/specid-{self.specimen_id}'
def __str__(self):
return (f'Specimen: {self.specimen_id} Peptide: {self.patient_peptide} Allele: {self.allele} Gene: {self.gene} ')
There's nothing stopping you declaring the TargetPrediction.gene field as a foreign key using the to_field attribute, so you wouldn't need to change the data at all:
class TargetPrediction(models.Model):
...
gene = models.ForeignKey("GeneTarget", to_field="gene")
Now your query simply becomes:
GeneTarget.objects.filter(list_name="whatever").values("gene").annotate(total=Count("targetprediction"))
I am trying to check if unique element is already present in postgresdb.
My method in views.py is
def bestfriend(username):
print username
user = Users.query.filter_by(username = username).first()
if user == None:
flash('bestfriend not found.')
return redirect(url_for('index'))
print user
u = g.user.friend(user)
#print bestfriend.id
if u is None:
#flash('Cannot be friend %(username)s.', username = username)
return redirect(url_for('user', username = username))
if db.session.query(bestfriend).filter(bestfriend.id==u.id).first():
flash('Already Exist')
return redirect(url_for('index'))
db.session.add(u)
db.session.commit()
flash('Your bestfriend has been added.')
return redirect(url_for('user', username = username))
My model.py is
bestfriend= db.Table('bestfriend',
db.Column('id',db.Integer, primary_key = True),
db.Column('friendid', db.Integer, db.ForeignKey('users.id'))
)
class Users(db.Model):
__tablename__ = "users"
id = db.Column(db.Integer, primary_key = True)
username = db.Column('username', db.String(20), unique=True , index=True)
password = db.Column('password' , db.String(10))
email = db.Column('email',db.String(50),unique=True , index=True)
registered_on = db.Column('registered_on' , db.DateTime)
posts = db.relationship('Post', backref = 'author', lazy = 'dynamic')
followed = db.relationship('Users',
secondary = followers,
primaryjoin = (followers.c.follower_id == id),
secondaryjoin = (followers.c.followed_id == id),
backref = db.backref('followers', lazy = 'dynamic'),
lazy = 'dynamic')
bestfriends = db.relationship('Users',
secondary = bestfriend,
primaryjoin = (bestfriend.c.friendid == id),
secondaryjoin = (bestfriend.c.id == id),
backref = db.backref('bestfriend', lazy = 'dynamic'),
lazy = 'dynamic')
I am able to insert value in database table bestfriend
Table "public.bestfriend"
Column | Type | Modifiers
----------+---------+-----------
id | integer | not null
friendid | integer |
Indexes:
"bestfriend_pkey" PRIMARY KEY, btree (id)
Foreign-key constraints:
"bestfriend_friendid_fkey" FOREIGN KEY (friendid) REFERENCES users(id)
but after insertion, I have to check if record is already present, using this line
if db.session.query(bestfriend).filter(bestfriend.id==u.id).first():
I am not sure about this statement. I have already tried similar solutions How to elegantly check the existence of an object/instance/variable and simultaneously assign it to variable if it exists in python? but it didn't worked for me.
I didn't receive any reply so adding some more info. I just want to check if id = 8 is already present in bestfriend table
app=> select * from bestfriend;
id | friendid
----+----------
8 | 11
Your statement does not work because bestfriend is a Table, not a mapped class. To fix the typo, just add c. to the query:
exists = db.session.query(bestfriend).filter(bestfriend.c.id==u.id).first()
Given that your relationship is defined as dynamic, you can do it easier with the following:
exists = user.bestfriends.filter(Users.id == u.id).one()
Side remark: Note that your bestfriend table is not structured correctly to allow many-to-many relationship. In order to do that, please change the definition of the table as per below:
bestfriend = db.Table('bestfriend',
db.Column('id',db.Integer, db.ForeignKey('users.id'), primary_key = True),
db.Column('friendid', db.Integer, db.ForeignKey('users.id'), primary_key = True)
)
Another remark: I do not understand the code flow: you add a friend at step u = g.user.friend(user), but later check if it exists. Shall logic not be reversed?
If you want to find if a record exists in a table, You can do it like this:
select 1 from table_name;
or better
select 1 from table_name where rownum=1;
The image below shows the query with output.
If there are any data, it will return 1 in the column.
Given this simple table written in SQLAlchemy and Django models.py, how would I set UPC to be unique if not null. UPC won't be available for all items, but if is it should be unique.
class Products(base):
__tablename__ = u'products'
id = Column(Integer(), primary_key=True, autoincrement = True)
product_name = Column(String(), unique=True, nullable=False)
upc = Column(String(), nullable = True)
And
class Products(models.Model):
id = models.AutoField(primary_key=True)
product_name = models.TextField()
upc = models.TextField(null=True, blank=True)
Multiple rows with NULL values should not be a problem for the unique constraint. Only "values" must be unique, NULL is no value.
Have you tried?:
upc = Column(String(), unique=True, nullable=True)
Background Schema:
class Checkpoint(db.Model):
id = db.Column(db.Integer, primary_key=True)
creator = db.Column(db.Integer, db.ForeignKey('user.id'))
name = db.Column(db.String(255))
description = db.Column(db.String(255), nullable=True)
price = db.Column(db.Float, nullable=True)
expiry = db.Column(db.DateTime, nullable=True)
date_created = db.Column(db.DateTime)
type = db.Column(db.String(255))
image = db.Column(db.String(255))
longitude = db.Column(db.Float)
latitude = db.Column(db.Float)
class UserCheckpoint(db.Model):
id = db.Column(db.Integer, primary_key=True)
user_id = db.Column(db.Integer, db.ForeignKey('user.id'))
user = db.relationship("User")
checkpoint_id = db.Column(db.Integer, db.ForeignKey('checkpoint.id'))
checkpoint = db.relationship("Checkpoint")
class User(db.Model):
id = db.Column(db.Integer, primary_key=True)
email = db.Column(db.String(255))
facebook_info = db.Column(db.String(255), db.ForeignKey('facebook_user.id'))
facebook_user = db.relationship("FacebookUser")
class FriendConnection(db.Model):
id = db.Column(db.Integer, primary_key=True)
fb_user_from = db.Column(db.String(255), db.ForeignKey('facebook_user.id'))
fb_user_to = db.Column(db.String(255), db.ForeignKey('facebook_user.id'))
class FacebookUser(db.Model):
id = db.Column(db.String(255), primary_key=True)
name = db.Column(db.String(255))
first_name = db.Column(db.String(255), nullable=True)
middle_name = db.Column(db.String(255), nullable=True)
last_name = db.Column(db.String(255), nullable=True)
gender = db.Column(db.String(255), nullable=True)
username = db.Column(db.String(255), nullable=True)
link = db.Column(db.String(255), nullable=True)
I have a user, and as you can see, each user has a Facebook profile, as well as a table depicting inter-facebook-profile friendships. So given the user, the user would have a list of Facebook friends. I would like to get all UserCheckpoints that belong either to the user or his friends, with a given Checkpoint condition:
coord_conditions = and_(Checkpoint.longitude <= longitude + exp_no,
Checkpoint.longitude >= longitude - exp_no,
Checkpoint.latitude <= latitude + exp_no,
Checkpoint.latitude >= latitude - exp_no,
)
How can I do this using the ORM from SQLAlchemy? Thanks!
Summary: How to select UserCheckpoints given that the user_id belong to a list of friends/self; while UserCheckpoint.checkpoint has a set of conditions to fulfill.
Each relation has two methods to defined conditions on related objects: .has() for single referred object and .any() for collections. These methods allow straightforward translation of your task to SQLAlchemy expression. Let's add missing relations to FacebookUser:
class FacebookUser(Model):
# Definitions from question are here
user = relationship(User, uselist=False)
friends = relationship('FacebookUser',
secondary=FriendConnection.__table__,
primaryjoin=(id==FriendConnection.fb_user_from),
secondaryjoin=(FriendConnection.fb_user_to==id))
I've defined FacebookUser.user assuming one-to-one relation (which is usually supplemented with unique constraint on the foreign key column). Just remove uselist=False and adjust name if you allow several users being connected to one facebook account.
A shorter definition of your condition for coordinates:
coord_conditions = Checkpoint.longitude.between(longitude - exp_no,
longitude + exp_no) & \
Checkpoint.latitude.between(latitude - exp_no,
latitude + exp_no)
This condition is definitely wrong even for approximation (-179.9° and 179.9° are very close, while the difference is huge), but this is not main topic of the question.
A condition for users of interest (user with id equal to user_id and his friends):
user_cond = (User.id==user_id) | \
User.facebook_user.has(
FacebookUser.friends.any(FacebookUser.user.has(id=user_id)))
Now the query is quite simple:
session.query(UserCheckpoint).filter(
UserCheckpoint.checkpoint.has(coord_conditions) & \
UserCheckpoint.user.has(user_cond))
Unless you have (or expect) performance issues, I'd suggest avoid optimizing it at the cost of readability.
Basically your query can be split in two parts:
Given the user_id, create a list of users which will contain the user herself as well as all direct friends
Given the list of users from 1., get all UserCheckpoint whose Checkpoint would satisfy the criteria.
Not tested code:
# get direct user for given user_id
u1 = (session.query(User.id.label("user_1_id"), User.id.label("user_id"))
)
# get friends of the user in one direction (from other user to this one)
User2 = aliased(User)
FacebookUser2 = aliased(FacebookUser)
u2 = (session.query(User2.id.label("user_1_id"), User.id.label("user_id")).
join(FacebookUser2, User2.facebook_info == FacebookUser2.id).
join(FriendConnection, FacebookUser2.id == FriendConnection.fb_user_from).
join(FacebookUser, FacebookUser.id == FriendConnection.fb_user_to).
join(User, User.facebook_info == FacebookUser.id)
)
# get friends of the user in other direction (from this user to the other)
User2 = aliased(User)
FacebookUser2 = aliased(FacebookUser)
u3 = (session.query(User2.id.label("user_1_id"), User.id.label("user_id")).
join(FacebookUser2, User2.facebook_info == FacebookUser2.id).
join(FriendConnection, FacebookUser2.id == FriendConnection.fb_user_to).
join(FacebookUser, FacebookUser.id == FriendConnection.fb_user_from).
join(User, User.facebook_info == FacebookUser.id)
)
# create a union to have all pairs (me_or_friend_id, user_id)
u_all = union_all(u1, u2, u3)
# **edit-1: added alias **
u_all = u_all.alias("user_list_view")
# final query which adds filters requirested (by user_id and the checkpoint condition)
q = (session.query(UserCheckpoint).
join(Checkpoint).filter(coord_conditions).
join(u_all, UserCheckpoint.user_id == u_all.c.user_1_id).
filter(u_all.c.user_id == user_id)
)
for u_cp in q.all():
print u_cp
Note, that you could simplify the query somewhat if you defined more relationships in your model and then can remove some primaryjoin conditions from join clauses.