How to use Python expression only as hybrid property in SQL Alchemy

Question

I have a SQL Alchemy mapping:

class Employee(Base):
  __tablename__ = 'employee'

  id = Column(Integer, primary_key=True)
  first_name = Column(String)
  last_name = Column(String)

  @hybrid_property
  def me(self):
    return 'Yes!' if self.first_name == 'ritratt' else 'Nope...'

Now I simply want to do a query like so: session.query(Employee.me).all()

But it does not work because SQL Alchemy expects a column or expression instead of a str. The solution is to use hybrid_expression but I do not want to use expressions. I want my mapping to only use hybrid property.

How can I write a mapping with a hybrid_property that returns a str and work with query()?

ritratt · Accepted Answer · 2019-09-17 16:50:39Z

1

I found a hacky way around it. I use Python to derive the value and then wrap it either in cast or concat:

from sqlalchemy import cast, String

...

@hybrid_property
  def me(self):
    value = 'Yes!' if self.first_name == 'ritratt' else 'Nope...'
    return cast(value, String) # or return concat('', value)

This still seems a bit hacky. Wondering if a better way exists

answered Sep 17, 2019 at 16:50

ritratt

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2 Comments

Andy Su Over a year ago

cast play good in this condition indeed ! another question is: can we use self.first_name to map a value ? So we can return sth like dict[self.first_name]

Andy Su Over a year ago

It seems self.first_name is not able to compare with string "ritratt"

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