How to check if value is already exist in database (SQLAlchemy) - sql

Question

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

Answer 1

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?

Answer 2

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.