SQLAlchemy IN clause

How about

session.query(MyUserClass).filter(MyUserClass.id.in_((123,456))).all()

edit: Without the ORM, it would be

session.execute(
    select(
        [MyUserTable.c.id, MyUserTable.c.name], 
        MyUserTable.c.id.in_((123, 456))
    )
).fetchall()

select() takes two parameters, the first one is a list of fields to retrieve, the second one is the where condition. You can access all fields on a table object via the c (or columns) property.

Assuming you use the declarative style (i.e. ORM classes), it is pretty easy:

query = db_session.query(User.id, User.name).filter(User.id.in_([123,456]))
results = query.all()

db_session is your database session here, while User is the ORM class with __tablename__ equal to "users".

An alternative way is using raw SQL mode with SQLAlchemy, I use SQLAlchemy 0.9.8, python 2.7, MySQL 5.X, and MySQL-Python as connector, in this case, a tuple is needed. My code listed below:

id_list = [1, 2, 3, 4, 5] # in most case we have an integer list or set
s = text('SELECT id, content FROM myTable WHERE id IN :id_list')
conn = engine.connect() # get a mysql connection
rs = conn.execute(s, id_list=tuple(id_list)).fetchall()

Hope everything works for you.

With the expression API, which based on the comments is what this question is asking for, you can use the in_ method of the relevant column.

To query

SELECT id, name FROM user WHERE id in (123,456)

use

myList = [123, 456]
select = sqlalchemy.sql.select([user_table.c.id, user_table.c.name], user_table.c.id.in_(myList))
result = conn.execute(select)
for row in result:
    process(row)

This assumes that user_table and conn have been defined appropriately.