带有标签并加入 postgres 的 sqlalchemy 核心答案

【问题标题】：sqlalchemy core with labels and joins postgres带有标签并加入 postgres 的 sqlalchemy 核心
【发布时间】：2017-01-20 20:27:39
【问题描述】：

使用 sqlalchemy 表达式语言，我希望从 2 个表中选择 id。因此我需要使用标签。如何解决这个问题。我需要相关表中的 id 和所有其他属性值。代码sn-p如下：

s1 = select([rt_issues.c.id.label('rt_issue_id'),
             rt_issues,
             queues.c.id.label('q_id'),
             queues,
            ]).\
    where(rt_issues.c.id == issue_id).\
    select_from(rt_issues.
    outerjoin(queues,
              rt_issues.c.queue_id == queues.c.id))
rs1 = conn.execute(s1)

错误日志：

“在 select 语句中尝试 'use_labels' 选项。” ％钥匙） InvalidRequestError：结果集中的列名“id”不明确！尝试 select 语句中的“use_labels”选项。

【问题讨论】：

标签： python postgresql sqlalchemy

【解决方案1】：

我用SQLAlchemy tutorial做了同样的用例。

>>> from sqlalchemy import Table, Column, Integer, String, MetaData, ForeignKey
>>> metadata = MetaData()
>>> users = Table('users', metadata,
...     Column('id', Integer, primary_key=True),
...     Column('name', String),
...     Column('fullname', String),
... )

>>> addresses = Table('addresses', metadata,
...   Column('id', Integer, primary_key=True),
...   Column('user_id', None, ForeignKey('users.id')),
...   Column('email_address', String, nullable=False)
...  )

所有数据

In [36]: s = select([users.c.id, users, addresses.c.id, addresses]).select_from(users.outerjoin(addresses))
In [37]: conn.execute(s).fetchall()
2017-01-20 19:13:11,218 INFO sqlalchemy.engine.base.Engine SELECT users.id, users.name, users.fullname, addresses.id, addresses.user_id, addresses.email_address
FROM users LEFT OUTER JOIN addresses ON users.id = addresses.user_id
2017-01-20 19:13:11,218 - sqlalchemy.engine.base.Engine - INFO - SELECT users.id, users.name, users.fullname, addresses.id, addresses.user_id, addresses.email_address
FROM users LEFT OUTER JOIN addresses ON users.id = addresses.user_id
2017-01-20 19:13:11,219 INFO sqlalchemy.engine.base.Engine {}
2017-01-20 19:13:11,219 - sqlalchemy.engine.base.Engine - INFO - {}
Out[37]:
[(1, 'jack', 'Jack Jones', 1, 1, 'jack@yahoo.com'),
 (1, 'jack', 'Jack Jones', 2, 1, 'jack@msn.com'),
 (2, 'wendy', 'Wendy Williams', 3, 2, 'www@www.org'),
 (2, 'wendy', 'Wendy Williams', 4, 2, 'wendy@aol.com')]

使用wherestatement

In [42]: s = select([users.c.id, users, addresses.c.id, addresses]).where(users.c.id == 1).select_from(users.outerjoin(addresses, addresses.c.user_id == users.c.id))
In [43]: conn.execute(s).fetchall()
2017-01-20 19:23:41,153 INFO sqlalchemy.engine.base.Engine SELECT users.id, users.name, users.fullname, addresses.id, addresses.user_id, addresses.email_address
FROM users LEFT OUTER JOIN addresses ON addresses.user_id = users.id
WHERE users.id = %(id_1)s
2017-01-20 19:23:41,153 - sqlalchemy.engine.base.Engine - INFO - SELECT users.id, users.name, users.fullname, addresses.id, addresses.user_id, addresses.email_address
FROM users LEFT OUTER JOIN addresses ON addresses.user_id = users.id
WHERE users.id = %(id_1)s
2017-01-20 19:23:41,155 INFO sqlalchemy.engine.base.Engine {'id_1': 1}
2017-01-20 19:23:41,155 - sqlalchemy.engine.base.Engine - INFO - {'id_1': 1}
Out[43]:
[(1, 'jack', 'Jack Jones', 1, 1, 'jack@yahoo.com'),
 (1, 'jack', 'Jack Jones', 2, 1, 'jack@msn.com')]

不需要标签

【讨论】：

它没有给你任何错误说明“使用标签”吗？我使用的是 sqlalchemy core 0.11。
@user956424 我使用SQLAlchemy==1.1.4 并且没有“标签”错误。 sqlalchemy core 0.11 是什么意思？
对不起，它应该被提到为 Flask v.0.11 和 sqlalchemy core v.1.0.8