database - Flask returning query -


i have model

class social(db.model): __tablename__ = 'social_auth_usersocialauth' id = db.column('id',db.integer, primary_key=true) provider = db.column('provider',db.string(32)) extra_data = db.column('extra_data',db.string()) uid = db.column('uid',db.string(255))  def __init__(self,id=none, provider=none, extra_data=none, uid=none):     self.id = id     self.provider = provider     self.extra_data = extra_data     self.uid = uid def __repr__(self):     return '<social %r>' % self.uid 

then here function

test = social.query.filter(social.uid == current_user) 

and when in on view {{ test }}

i it's query, want result , how can ? here result on view now:

select social_auth_usersocialauth.id social_auth_usersocialauth_id, social_auth_usersocialauth.provider social_auth_usersocialauth_provider, social_auth_usersocialauth.extra_data social_auth_usersocialauth_extra_data, social_auth_usersocialauth.uid social_auth_usersocialauth_uid social_auth_usersocialauth social_auth_usersocialauth.uid = :uid_1 

you have 2 problems. let's start error.

in query, compare uid field current user. assuming social user model, query should include

social.query.filter(social.uid == current_user.uid) 

your other problem how access results of query. methods filter , filter_by return basequery(http://flask-sqlalchemy.pocoo.org/2.1/api/#flask.ext.sqlalchemy.basequery). basequery objects have several methods expose results depending on needs. in case sounds want all(http://flask-sqlalchemy.pocoo.org/2.1/api/#flask.ext.sqlalchemy.basequery.all).

putting get

test = social.query.filter(social.uid == current_user.uid).all() 

once in template, you'll want iterate on test rather echoing directly.

{% user in test %}   {{ user }} {% endfor %} 

if haven't defined social.__str__, output won't useful, should enough moving in right direction.