在SQLAlchemy中选择Null
||
我想做相当于
SELECT * FROM
(SELECT foo, bar FROM baz JOIN quux ON baz.id = quux.id
UNION
SELECT foo, NULL AS bar FROM baz)
GROUP BY (foo, bar) HAVING foo = \'John Doe\';
NULL
q1 = session.query(Baz.foo, Quux.bar).join(Quux)
q2 = session.query(Baz.foo, None)
# ^^^^ This breaks!
没有找到相关结果
已邀请:
3 个回复
抢垢洛韧
import sqlalchemy as sa from sqlalchemy.orm import sessionmaker from sqlalchemy.ext.declarative import declarative_base mymetadata = sa.MetaData() Base = declarative_base(metadata=mymetadata) Session = sessionmaker(bind=sa.engine) session = Session() class Person(Base): __tablename__ = \'some_table\' id = sa.Column(sa.Integer, primary_key=True) name = sa.Column(sa.String(50)) print sa.select([Person.name, sa.text(\'NULL as null_bar\')]) >>> SELECT some_table.name, NULL as null_bar FROM some_table妊辽剁茧
q1 = session.query(Baz.foo, Quux.bar) \\ .join(Quux.bar) q2 = session.query(Baz.foo, sqlalchemy.null().label(\'null_bar\')) qall = q1.union(q2) foocol = qall.column_descriptions[0][\'expr\'] qgrp = qall.group_by([col[\'name\'] for col in qall.column_descriptions]) q = qgrp.having(foocol == \'John Doe\') q.all()笛驮型迸
q1 = session.query(Baz.foo, Quux.bar) \\ .join(Quux.bar) q2 = session.query(Baz.foo, sqlalchemy.sql.expression.literal_column(\'NULL as null_bar\')) qall = q1.union(q2) foocol = qall.column_descriptions[0][\'expr\'] qgrp = qall.group_by([col[\'name\'] for col in qall.column_descriptions]) q = qgrp.having(foocol == \'John Doe\') q.all()