Sqlalchemy：FROM 中的子查询必须有一个别名

Question

我如何构建这个 sqlalchemy 查询以使其做正确的事情？

我已经给出了所有我能想到的别名，但我仍然得到：

ProgrammingError: (psycopg2.ProgrammingError) subquery in FROM must have an alias
LINE 4: FROM (SELECT foo.id AS foo_id, foo.version AS ...

此外，正如 IMSoP 指出的那样，它似乎试图将其变成交叉连接，但我只是希望它通过子查询在同一个 [=48= 上加入 table 和组].

这是sql炼金术：

（注意：我已将其重写为一个尽可能完整的独立文件，并且可以运行来自 python shell）

from sqlalchemy import create_engine, func, select
from sqlalchemy import Column, BigInteger, DateTime, Integer, String, SmallInteger
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

engine = create_engine('postgresql://postgres:#######@localhost:5435/foo1234')
session = sessionmaker()
session.configure(bind=engine)
session = session()

Base = declarative_base()

class Foo(Base):
     __tablename__ = 'foo'
     __table_args__ = {'schema': 'public'}
     id = Column('id', BigInteger, primary_key=True)
     time = Column('time', DateTime(timezone=True))
     version = Column('version', String)
     revision = Column('revision', SmallInteger)

foo_max_time_q = select([
     func.max(Foo.time).label('foo_max_time'),
     Foo.id.label('foo_id')
 ]).group_by(Foo.id
 ).alias('foo_max_time_q')

foo_q = select([
    Foo.id.label('foo_id'),
    Foo.version.label('foo_version'),
    Foo.revision.label('foo_revision'),
    foo_max_time_q.c.foo_max_time.label('foo_max_time')
]).join(foo_max_time_q, foo_max_time_q.c.foo_id == Foo.id
).alias('foo_q')

thing = session.query(foo_q).all()
print thing

生成sql:

SELECT foo_id AS foo_id,
    foo_version AS foo_version,
    foo_revision AS foo_revision,
    foo_max_time AS foo_max_time,
    foo_max_time_q.foo_max_time AS foo_max_time_q_foo_max_time,
    foo_max_time_q.foo_id AS foo_max_time_q_foo_id
    FROM (SELECT id AS foo_id,
        version AS foo_version,
        revision AS foo_revision,
        foo_max_time_q.foo_max_time AS foo_max_time
        FROM (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
        ) AS foo_max_time_q)
    JOIN (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
    ) AS foo_max_time_q
    ON foo_max_time_q.foo_id = id

和这是玩具table:

CREATE TABLE foo (
id bigint ,
time timestamp with time zone,
version character varying(32),
revision smallint
);

我期望得到的SQL（期望SQL）是这样的：

SELECT foo.id AS foo_id,
       foo.version AS foo_version,
       foo.revision AS foo_revision,
       foo_max_time_q.foo_max_time AS foo_max_time
       FROM foo
       JOIN (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
            ) AS foo_max_time_q
        ON foo_max_time_q.foo_id = foo.id

最后的注释：如果可能的话，我希望使用 select() 而不是 session.query() 得到答案。谢谢

Answer 1

你快到了。做一个"selectable" subquery and join it with the main query via join():

foo_max_time_q = select([func.max(Foo.time).label('foo_max_time'),
                         Foo.id.label('foo_id')
                        ]).group_by(Foo.id
                         ).alias("foo_max_time_q")

foo_q = session.query(
          Foo.id.label('foo_id'),
          Foo.version.label('foo_version'),
          Foo.revision.label('foo_revision'),
          foo_max_time_q.c.foo_max_time.label('foo_max_time')
                ).join(foo_max_time_q, 
                       foo_max_time_q.c.foo_id == Foo.id)

print(foo_q.__str__())

打印（手动美化）：

SELECT 
    foo.id AS foo_id, 
    foo.version AS foo_version, 
    foo.revision AS foo_revision, 
    foo_max_time_q.foo_max_time AS foo_max_time 
FROM 
    foo 
JOIN 
    (SELECT 
         max(foo.time) AS foo_max_time, 
         foo.id AS foo_id 
     FROM 
         foo 
     GROUP BY foo.id) AS foo_max_time_q 
ON 
    foo_max_time_q.foo_id = foo.id

此 gist 中提供了完整的工作代码。

Answer 2

原因

subquery in FROM must have an alias

这个错误意味着子查询（我们试图对其执行join）没有别名。
即使我们 .alias('t') 它只是为了满足这个要求，我们也会得到下一个错误：

missing FROM-clause entry for table "foo"

那是因为join on子句（... == Foo.id）不熟悉Foo。
它只知道“左”和“右”表：t（子查询）和foo_max_time_q.

解决方案

相反，select_from Foo 和 foo_max_time_q 的连接。

方法一

将.join(B, on_clause)替换为.select_from(B.join(A, on_clause):

]).join(foo_max_time_q, foo_max_time_q.c.foo_id == Foo.id

]).select_from(foo_max_time_q.join(Foo, foo_max_time_q.c.foo_id == Foo.id)

这在这里有效，因为 A INNER JOIN B 等同于 B INNER JOIN A。

方法二

要保留联接表的顺序：

from sqlalchemy import join

并将 .join(B, on_clause) 替换为 .select_from(join(A, B, on_clause)):

]).join(foo_max_time_q, foo_max_time_q.c.foo_id == Foo.id

]).select_from(join(Foo, foo_max_time_q, foo_max_time_q.c.foo_id == Foo.id)

可以找到 session.query() 的替代方法。

Sqlalchemy：FROM 中的子查询必须有一个别名

Sqlalchemy: subquery in FROM must have an alias

python

postgresql

select

sqlalchemy

psycopg2

原因

解决方案

方法一

方法二