Hi Mike, Yes the SQL code runs as desired, but the Python code doesn't, unfortunately. After reading the references you pointed out, my Python code looks like this:
class Staff(Base): id = Column(Integer, primary_key=True) selfreferencing_staff_id = Column( Integer, ForeignKey('staff_table.id', onupdate="CASCADE", ondelete='SET NULL')) _enabling_factor = Column(Integer) team_members = relationship('Staff', backref=backref( 'supervisor', remote_side='Staff.id')) anchor_s = Staff.__table__.alias(name="anchor_s") s = Staff.__table__.alias(name="s") final_s = Staff.__table__.alias(name="final_s") recursive_cte = select([ anchor_s.c.id, anchor_s.c._enabling_factor, anchor_s.c.selfreferencing_staff_id ]).select_from(anchor_s).where(anchor_s.c.id == Staff.id).cte( name="lineage_nodes", recursive=True) lineage_nodes = recursive_cte.union_all( select([ s.c.id, s.c._enabling_factor, s.c.selfreferencing_staff_id ]).join(recursive_cte, recursive_cte.c.selfreferencing_staff_id == s.c.id).filter(recursive_cte.c._enabling_factor == None)) top_id = select(sasql.func.min(lineage_nodes.c.id)) Staff.effective_enabling_factor = column_property(select(final_s.c._enabling_factor).where( final_s.c.record_id == top_id).scalar_subquery()) # This is where I define the desired column_property. The problems seems to be that this code doesn't generate `correlated subquery` for each `Staff` row ( as in the SQL code `WHERE anchor_s.id = outer_s.id`). Could you take a look? On Saturday, September 25, 2021 at 8:26:46 PM UTC-7 Mike Bayer wrote: > well __class__.__table__ isn't going to be there inside the class body, > just to make things simple you will want to add this column_property() > after the Staff class is fully defined; then you make use of Staff.<col> to > get at columns. > https://docs.sqlalchemy.org/en/14/orm/mapped_sql_expr.html#using-column-property > > will show how to assign a new column_property to the class after the fact. > > next, the query you have is correlating, but i really dont have any idea > if SQLAlchemy is going to handle a correlated column deep inside of a CTE > like that. The form you have, with the CTE embedded in the parenthesis, > is only possible with the most recent SQLAlchemy 1.4.24 release where > someone contributed a new parameter called "nesting", which means the CTE > should not be moved to the top level of the SELECT. still, not really > sure what a CTE will do inside a correlated subquery like that. For an > example of how to use "nesting" see example four at > https://docs.sqlalchemy.org/en/14/core/selectable.html?highlight=hascte#sqlalchemy.sql.expression.HasCTE.cte > > . for the correlate, when you write out the select() that refers to > "outer_s", add .correlate(Staff) to it, which means Staff isn't added to > the FROM list, it's assumed to be on the outside. > > give those a try but im not totally sure CTEs work as correlated > subqueries right now, it's not been tried. I assume you've confirmed this > query actually runs, im surprised you can even correlate inside of a CTE > like that. > > > > > > On Sat, Sep 25, 2021, at 12:04 AM, niuji...@gmail.com wrote: > > Hi Mike, thanks for pointing out the direction. > I've worked out the SQL, but failed when converting to SQLAlchemy > construct. > > My SQL query looks like this: > > SELECT id, ( > WITH lineage_nodes (id, _enabling_factor, > selfreferencing_staff_id) AS > ( > SELECT anchor_s.id, anchor_s._enabling_factor, > anchor_s.selfreferencing_staff_id > FROM staff_table AS anchor_s > WHERE anchor_s.id = outer_s.id > > UNION ALL > > SELECT s.id, s._enabling_factor, s.selfreferencing_staff_id > FROM lineage_nodes AS l > INNER JOIN staff_table AS s > ON l.selfreferencing_staff_id = s.id > WHERE l._enabling_factor IS NULL > ), > > top_node_id (top_id) AS > ( > SELECT MIN(id) AS top_id FROM lineage_nodes > ) > > SELECT staff_table._enabling_factor > FROM staff_table > INNER JOIN top_node_id > ON staff_table.id = top_node_id.top_id > ) AS effective_enabling_factor > FROM staff_table AS outer_s; > > > > > My Python codes looks like this: > > > class Staff(Base): > id = Column(Integer, primary_key=True) > selfreferencing_staff_id = Column( > Integer, > ForeignKey('staff_table.id', > onupdate="CASCADE", > ondelete='SET NULL')) > _enabling_factor = Column(Integer) > > ## codes below doesn't work: > anchor_s = __class__.__table__.alias(name="anchor_s") > s = __class__.__table__.alias(name="s") > > recursive_cte = select([ > id, _enabling_factor, selfreferencing_staff_id > ]).select_from(anchor_s).where(anchor_s.c.id == id).cte( > name="lineage_nodes", recursive=True) > > lineage_nodes = recursive_cte.union_all( > select([ > s.c.id, s.c._enabling_factor, s.c.selfreferencing_staff_id > ]).join(recursive_cte, > recursive_cte.c.selfreferencing_staff_id == > s.c.id).filter(recursive_cte.c._enabling_factor == None)) > > top_id = select(sasql.func.min(lineage_nodes.c.id)) > effective_enabling_factor = column_property(...) # I have trouble > in this line here. > ## codes above has a NameError: name '__class__' is not defined > > team_members = relationship('Staff', > backref=backref( > 'supervisor', > remote_side='Staff.id')) > On Friday, September 24, 2021 at 1:00:01 PM UTC-7 Mike Bayer wrote: > > > this is a hefty query to dig in to but column_property() subqueries have > to be formed in terms of a correlated subquery. So instead of injecting a > particular primary key into it, you set it to point to the Staff.id column. > > correlated subqueries are not terrific performers and the construct can be > a little bit clumsy in the ORM as well, however, the second example at > https://docs.sqlalchemy.org/en/14/orm/mapped_sql_expr.html#using-column-property > > shows the general idea. > > a first step to understanding might be to write out the SQL you think you > want when you SELECT some Staff rows, where one of the columns in the row > is the "effective_enabling_factor". that column needs to be written as a > correlated subquery for it to be compatible with column_property(). > > > > On Fri, Sep 24, 2021, at 1:47 AM, niuji...@gmail.com wrote: > > class Staff(Base): > id = Column(Integer, primary_key=True) > selfreferencing_staff_id = Column( > Integer, > ForeignKey('staff_table.id', > onupdate="CASCADE", > ondelete='SET NULL')) > _enabling_factor = Column(Integer) > effective_enabling_factor = column_property(...) # I have trouble in > this line here. > team_members = relationship('Staff', > backref=backref( > 'supervisor', > remote_side='Staff.id')) > > > This is a self-referencing lineage. Each staff has one supervisor above > them. Each staff has a `_enabling_factor`, which can be either a Integer, > or Null. A staff's `effective_enabling_factor` is either their own > `_enabling_factor` value, or their supervisor's `effective_enabling_factor` > if their own is Null. > > This seems to be a case to use recursive CTE. > I can construct the query for a certain staff member, e.g. staff #5: > > recursive_cte = select([Staff.id, Staff._enabling_factor, > Staff.selfreferencing_staff_id]).where(Staff.id==5).cte(recursive=True) > > lineage_nodes = recursive_cte.union_all(select([Staff.id, > Staff._enabling_factor, > Staff.selfreferencing_staff_id]).join(recursive_cte, > recursive_cte.c.selfreferencing_staff_id==Staff.record_id).filter(recursive_cte.c._enabling_factor > > == None)) > > marker_carrying_supervisor_id = select(sasql.func.min(lineage_nodes.c.id > )).scalar_subquery() > > > select(Staff._enabling_factor).where(Staff.id==marker_carrying_supervisor_id) > > > However, I don't see how I can write this recursive CTE as a > column_property on the `Staff` class. Instead of giving specific primary > key (e.g. #5), I need to somehow reference current row as the anchor. > > How to solve this? > > > -- > SQLAlchemy - > The Python SQL Toolkit and Object Relational Mapper > > http://www.sqlalchemy.org/ > > To post example code, please provide an MCVE: Minimal, Complete, and > Verifiable Example. See http://stackoverflow.com/help/mcve for a full > description. > --- > You received this message because you are subscribed to the Google Groups > "sqlalchemy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sqlalchemy+...@googlegroups.com. > To view this discussion on the web visit > https://groups.google.com/d/msgid/sqlalchemy/a5e26719-3e5e-4f6e-92fc-c3319ad3ec6fn%40googlegroups.com > > <https://groups.google.com/d/msgid/sqlalchemy/a5e26719-3e5e-4f6e-92fc-c3319ad3ec6fn%40googlegroups.com?utm_medium=email&utm_source=footer> > . > > > > -- > SQLAlchemy - > The Python SQL Toolkit and Object Relational Mapper > > http://www.sqlalchemy.org/ > > To post example code, please provide an MCVE: Minimal, Complete, and > Verifiable Example. 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