Question

我正在尝试将表中的几个字段分组，然后对这些组进行求和，但它们会被重复计算。

我的模型如下：

class CostCenter(db.Model):
     __tablename__ = 'costcenter'
     id = db.Column(db.Integer, primary_key=True, autoincrement=True)
     name = db.Column(db.String)
     number = db.Column(db.Integer)

class Expense(db.Model):

    __tablename__ = 'expense'
    id = db.Column(db.Integer, primary_key=True, autoincrement=True)
    glitem_id = db.Column(db.Integer, db.ForeignKey('glitem.id'))
    glitem = db.relationship('GlItem')
    costcenter_id = db.Column(db.Integer, db.ForeignKey('costcenter.id'))
    costcenter = db.relationship('CostCenter')
    value = db.Column(db.Float)
    date = db.Column(db.Date)

我一直在使用：

expenses=db.session.query(Expense,func.sum(Expense.value)).group_by(Expense.date).filter(CostCenter.id.in_([1,2,3]))

当我打印费用时，它显示了后面的SQL语句。它对我来说是正确的，但我不熟悉SQL。问题是它作为sum_1输出的值被多次计数。如果我在＆＃34;中有[1]项，则在＃34;它将三者相加。如果我有[1,2]，它将总和所有三个然后加倍它，如果我有[1,2,3]，它将总和所有三个并加倍它。我不确定为什么它会多次计数。我该如何解决？

SELECT expense.id AS expense_id, expense.glitem_id AS expense_glitem_id, expense.costcenter_id AS         expense_costcenter_id, expense.value AS expense_value, expense.date AS expense_date, sum(expense.value) AS sum_1 
FROM expense, costcenter 
WHERE costcenter.id IN (:id_1, :id_2, :id_3) GROUP BY expense.date

谢谢！

Answer 1

这里有一些问题;你似乎不是在追问正确的事情。按Expense.date分组时选择Expense对象毫无意义。 CostCenter和Expense之间需要有一些连接条件，否则行将被复制，每个成本中心都计入，但两者之间没有关系。

您的查询应如下所示：

session.query(
    Expense.date,
    func.sum(Expense.value).label('total')
).join(Expense.cost_center
).filter(CostCenter.id.in_([2, 3])
).group_by(Expense.date
).all()

生成这个sql：

SELECT expense.date AS expense_date, sum(expense.value) AS total 
FROM expense JOIN cost_center ON cost_center.id = expense.cost_center_id 
WHERE cost_center.id IN (?, ?) GROUP BY expense.date

这是一个简单的可运行示例：

from datetime import datetime
from sqlalchemy import create_engine, Column, Integer, ForeignKey, Numeric, DateTime, func
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import Session, relationship

engine = create_engine('sqlite://', echo=True)
session = Session(bind=engine)
Base = declarative_base(bind=engine)


class CostCenter(Base):
    __tablename__ = 'cost_center'

    id = Column(Integer, primary_key=True)


class Expense(Base):
    __tablename__ = 'expense'

    id = Column(Integer, primary_key=True)
    cost_center_id = Column(Integer, ForeignKey(CostCenter.id), nullable=False)
    value = Column(Numeric(8, 2), nullable=False, default=0)
    date = Column(DateTime, nullable=False)

    cost_center = relationship(CostCenter, backref='expenses')


Base.metadata.create_all()

session.add_all([
    CostCenter(expenses=[
        Expense(value=10, date=datetime(2014, 8, 1)),
        Expense(value=20, date=datetime(2014, 8, 1)),
        Expense(value=15, date=datetime(2014, 9, 1)),
    ]),
    CostCenter(expenses=[
        Expense(value=45, date=datetime(2014, 8, 1)),
        Expense(value=40, date=datetime(2014, 9, 1)),
        Expense(value=40, date=datetime(2014, 9, 1)),
    ]),
    CostCenter(expenses=[
        Expense(value=42, date=datetime(2014, 7, 1)),
    ]),
])
session.commit()

base_query = session.query(
    Expense.date,
    func.sum(Expense.value).label('total')
).join(Expense.cost_center
).group_by(Expense.date)

# first query considers center 1, output:
# 2014-08-01: 30.00
# 2014-09-01: 15.00
for row in base_query.filter(CostCenter.id.in_([1])).all():
    print('{}: {}'.format(row.date.date(), row.total))

# second query considers centers 1, 2, and 3, output:
# 2014-07-01: 42.00
# 2014-08-01: 75.00
# 2014-09-01: 95.00
for row in base_query.filter(CostCenter.id.in_([1, 2, 3])).all():
    print('{}: {}'.format(row.date.date(), row.total))

SQLAlchemy中的GroupBy和Sum？

1 个答案: