1

我正在尝试获取每个团队的最新变化。

SQLite 语句
按预期工作。

SELECT * FROM (
  SELECT * FROM team_history ORDER BY changed_at DESC
) sub GROUP BY team

SQLAlchemy 实现
无论出于何种原因，我必须使用排序asc()而不是desc()获得相同的结果，这就是为什么我怀疑我的实现是否正确。

session.query(TeamHistory)\
    .select_entity_from(
        session.query(TeamHistory).order_by(asc(TeamHistory.changed_at)).subquery()
    ).group_by(TeamHistory.team)\
    .all()

环境

Python：3.8.0
SQLAlchemy：1.3.23

复制

架构：

CREATE TABLE "team_history" (ID integer PRIMARY KEY, changed_at TEXT, team TEXT);

记录：

[{"ID":1,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":2,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":3,"changed_at":"2021-03-02 10:30:00","team":"B"},
 {"ID":4,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":5,"changed_at":"2021-03-02 11:30:00","team":"B"},
 {"ID":6,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":7,"changed_at":"2021-03-02 11:00:00","team":"B"},
 {"ID":8,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":9,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":10,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":11,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":12,"changed_at":"2021-03-02 10:00:00","team":"A"},
 {"ID":13,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":14,"changed_at":"2021-03-02 12:30:00","team":"A"},
 {"ID":15,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":16,"changed_at":"2021-03-02 12:00:00","team":"A"},
 {"ID":17,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":18,"changed_at":"2021-03-02 13:30:00","team":"A"},
 {"ID":19,"changed_at":"2021-03-02 10:00:00","team":"B"},
 {"ID":20,"changed_at":"2021-03-02 10:00:00","team":"A"}]

解决
方法 谢谢大家！

session.query(TeamHistory)\
    .group_by(TeamHistory.team)\
    .having(func.max(TeamHistory.changed_at))\
    .all()
4

2 回答 2

2

当您使用：

SELECT *
FROM tablename
GROUP BY somecolumn

SQLite 为 的每个不同值返回 1 行somecolumn，但哪一行？
该文档指出该行未定义，这意味着它是任意选择的，尽管根据我的经验，似乎将返回结果集中属于每个组的第一行。
但这并不能保证，并且应该避免上述查询和您的查询。

有多种方法可以获取team最新的每一行changed_at。
其中一个在 SQLite 中有效（尽管它在其他数据库中不起作用）是：

SELECT * FROM team_history GROUP BY team HAVING MAX(changed_at)

见演示。
因此，这是您应该转换为 SQLAlchemy 的查询（我对此无能为力）。

还有其他方法，使用窗口函数或EXISTS.

于 2021-03-02T16:44:26.840 回答
1

基本上，您提出的查询应该可以工作。SQLite 将日期存储为 ISO-8601 格式的字符串，该格式具有字典顺序和时间顺序相同的属性。同样使用 datetime 列TEXT具有相同的属性。

因此，子查询的排序desc应导致以下结果：

ID 约会时间 团队
18 2021-03-02 13:30:00 一个
14 2021-03-02 12:30:00 一个
16 2021-03-02 12:00:00 一个
5 2021-03-02 11:30:00 乙
.. .. ..

如另一个答案所述，group by 的问题是任意选择每个组的返回行，但看起来它始终是第一行。知道了这一点，我们有不同的解决方案来确定它应该返回哪些行：

# Using an aggregate function in select
session.query(TeamHistory.team, func.max(TeamHistory.changed_at)).group_by(TeamHistory.team).all()

# Using an aggregate function with `having`
session.query(TeamHistory).group_by(TeamHistory.team).having(func.max(TeamHistory.changed_at).all()

这将产生以下工作示例：

from sqlalchemy import Column, Integer, create_engine, Text, DateTime, func
from sqlalchemy.orm import sessionmaker, deferred, column_property
from sqlalchemy.ext.declarative import declarative_base
import datetime

Base = declarative_base()

class TeamHistory(Base):
    __tablename__ = 'team_history'
    id = Column(Integer, primary_key=True)
    changed_at = Column(DateTime)
    team = Column(Text)

if __name__ == '__main__':
    engine = create_engine('sqlite://')
    Base.metadata.create_all(engine)
    Session = sessionmaker(engine)

    db = Session()
    
    lst = [{"ID":1,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":2,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":3,"changed_at":"2021-03-02 10:30:00","team":"B"},{"ID":4,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":5,"changed_at":"2021-03-02 11:30:00","team":"B"},{"ID":6,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":7,"changed_at":"2021-03-02 11:00:00","team":"B"},{"ID":8,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":9,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":10,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":11,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":12,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":13,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":14,"changed_at":"2021-03-02 12:30:00","team":"A"},{"ID":15,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":16,"changed_at":"2021-03-02 12:00:00","team":"A"},{"ID":17,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":18,"changed_at":"2021-03-02 13:30:00","team":"A"},{"ID":19,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":20,"changed_at":"2021-03-02 10:00:00","team":"A"}]
    for dct in lst:
        dt = datetime.datetime.strptime(dct.get('changed_at'), '%Y-%m-%d %H:%M:%S')
        nth = TeamHistory(id=dct.get('ID'), changed_at=dt, team=dct.get('team'))
        db.add(nth)

    db.commit()

    res = db.query(TeamHistory)\
        .select_entity_from(
            db.query(TeamHistory).order_by(TeamHistory.changed_at.desc()).subquery()
        ).group_by(TeamHistory.team)\
        .all()

    for r in res:
        print(r.id, r.changed_at, r.team)
    print()


    res = db.query(TeamHistory)\
        .group_by(TeamHistory.team)\
        .having(func.max(TeamHistory.changed_at))\
        .all()

    for r in res:
        print(r.id, r.changed_at, r.team)
    print()

    res = db.query(TeamHistory.id, func.max(TeamHistory.changed_at), TeamHistory.team)\
        .group_by(TeamHistory.team)\
        .all()

    for r in res:
        print(r[0], r[1], r[2])

如果您主要对团队的最后一次更改感兴趣，您也可以使用deferredand column_propertyto only get MAX(changed_at)。

from sqlalchemy import Column, Integer, create_engine, Text, DateTime, func, select
from sqlalchemy.orm import sessionmaker, deferred, column_property
from sqlalchemy.ext.declarative import declarative_base
import datetime

Base = declarative_base()

class TeamHistory(Base):
    __tablename__ = 'team_history_def'
    id = Column(Integer, primary_key=True)
    changed_at = deferred(Column(DateTime))
    last_changed = column_property(func.max(changed_at))
    team = Column(Text)

if __name__ == '__main__':
    engine = create_engine('sqlite://')
    Base.metadata.create_all(engine)
    Session = sessionmaker(engine)

    db = Session()
    
    lst = [{"ID":1,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":2,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":3,"changed_at":"2021-03-02 10:30:00","team":"B"},{"ID":4,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":5,"changed_at":"2021-03-02 11:30:00","team":"B"},{"ID":6,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":7,"changed_at":"2021-03-02 11:00:00","team":"B"},{"ID":8,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":9,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":10,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":11,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":12,"changed_at":"2021-03-02 10:00:00","team":"A"},{"ID":13,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":14,"changed_at":"2021-03-02 12:30:00","team":"A"},{"ID":15,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":16,"changed_at":"2021-03-02 12:00:00","team":"A"},{"ID":17,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":18,"changed_at":"2021-03-02 13:30:00","team":"A"},{"ID":19,"changed_at":"2021-03-02 10:00:00","team":"B"},{"ID":20,"changed_at":"2021-03-02 10:00:00","team":"A"}]
    for dct in lst:
        dt = datetime.datetime.strptime(dct.get('changed_at'), '%Y-%m-%d %H:%M:%S')
        nth = TeamHistory(id=dct.get('ID'), changed_at=dt, team=dct.get('team'))
        db.add(nth)

    db.commit()

    res = db.query(TeamHistory)\
        .group_by(TeamHistory.team)\
        .all()

    for r in res:
        print(r.id, r.changed_at, r.team)
    print()
于 2021-03-05T09:24:52.890 回答