Manlio Perillo ha scritto:
[...]
op = (a.c.x / a.c.y).label('z')
query = select([op], order_by=[op], use_labels=True)
The solution is quite simple (and very elegant):
s = select([op], use_labels=True)
s.order_by(s.c.z)
Regards Manlio Perillo
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