Sqlalchemy get last X rows in order

Question

I need to get the last X rows from a table, but in order of the ID. How could I achieve this?

Answer 1

query = users.select().order_by(users.c.id.desc()).limit(5)  print reversed(conn.execute(query).fetchall() ) 

something like that anyway

Answer 2

This worked for me...

c=session.query(a).order_by(a.id.desc()).limit(2)
c=c[::-1]

This solution is 10 times faster than the python slicing solution proposed by BrendanSimon.

Answer 3

I believe you can prefix the order_by parameter with a '-' to get reverse order.

query = users.select().order_by(-users.c.id.desc()).limit(5)

Also, I believe you can use python slices as an alternative to limit.

query = users.select().order_by(users.c.id.desc())[-5:]

query = users.select().order_by(-users.c.id.desc())[:5]