Multi Database with same table name in flask_sqlalchemy - sqlalchemy

Question

I have multi databases, like order_db1、order_db2、order_db3,
but they all have table order, but data in order_db1.order is different from order_db2.order,
order_db1.order
| id | amount|
| 1| 100|
order_db2.order
| id | amount|
| 1| 200|
how can I access multi database with same table name in flask_sqlalchemy ？
I tried like this，use SQLALCHEMY_BINDS：
# -*- coding: utf-8 -*-
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
import logging
class Application(Flask):
def __init__(self, import_name):
super(Application, self).__init__(import_name)
app = Application(__name__)
db = SQLAlchemy(app)
app.config['SQLALCHEMY_BINDS'] = {
'order_db1': 'mysql://%s:%s#%s/order_db1' % (mysql_user, quote_plus(mysql_pwd),mysql_host),
'order_db2': 'mysql://%s:%s#%s/order_db2' % (mysql_user, quote_plus(mysql_pwd),mysql_host),
}
class Order1(db.model):
__bind_key__ = 'order_db1'
__tablename__ = 'order'
id = db.colum(db.Interger)
amount = db.colum(db.Interger)
class Order2(db.model):
__bind_key__ = 'order_db2'
__tablename__ = 'order'
id = db.colum(db.Interger)
amount = db.colum(db.Interger)
q1=Order1.query.get(1)
q2=Order2.query.get(1)
It told me :
'sqlalchemy.exc.InvalidRequestError: Table 'order' is already defined for this MetaData instance. Specify 'extend_existing=True' to redefine options and columns on an existing Table object.'
But if I add __table_args = {'extend_existing': True}
class Order1(db.model):
__bind_key__ = 'order_db1'
__tablename__ = 'order'
__table_args = {'extend_existing': True}
id = db.colum(db.Interger)
amount = db.colum(db.Interger)
class Order2(db.model):
__bind_key__ = 'order_db2'
__tablename__ = 'order'
__table_args = {'extend_existing': True}
id = db.colum(db.Interger)
amount = db.colum(db.Interger)
q1=Order1.query.get(1)
q2=Order2.query.get(1)
the data in q1 is the same with q2, means id=1,amount=200
I want it like :
q1: id=1, amount=100
q2: id=1, amount=200

Answer 1

There are known issues with binds and identical table names. See this discussion and this possible workaround.
An alternative workaround is to use two Sqlalchemy instances, which may be or not acceptable for your use case. See single file example using SQLite.
import random
from flask import Flask, render_template_string
from flask_sqlalchemy import SQLAlchemy
app = Flask(__name__)
app.config['SQLALCHEMY_BINDS'] = {
'order_db1': 'sqlite:///order_db1.sqlite',
'order_db2': 'sqlite:///order_db2.sqlite',
}
db1 = SQLAlchemy(app)
db2 = SQLAlchemy(app)
class Order1(db1.Model):
__bind_key__ = 'order_db1'
__tablename__ = 'order'
id = db1.Column(db1.Integer, primary_key=True)
amount = db1.Column(db1.Integer)
class Order2(db2.Model):
__bind_key__ = 'order_db2'
__tablename__ = 'order'
id = db2.Column(db2.Integer, primary_key=True)
amount = db2.Column(db2.Integer)
_html_template = '''
<p>Order 1 count:{{order_1_count}}</p>
<p>Order 2 count:{{order_2_count}}</p>
'''
#app.route('/')
def index():
order_1_count = Order1.query.count()
order_2_count = Order2.query.count()
return render_template_string(_html_template, order_1_count=order_1_count, order_2_count=order_2_count)
#app.before_first_request
def build_sample_db():
db1.drop_all()
db2.drop_all()
db1.create_all()
db2.create_all()
# Create 100 records in db1
db1.session.add_all([Order1(amount=random.randint(0, 100)) for _ in enumerate(range(0, 100))])
db1.session.commit()
# Create 1000 records in db1
db2.session.add_all([Order2(amount=random.randint(1000, 10000)) for _ in enumerate(range(0, 1000))])
db2.session.commit()
return app
if __name__ == '__main__':
app.run()