SQL炼金术中的GroupBy和Sum? [英] GroupBy and Sum in SQLAlchemy?

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本文介绍了SQL炼金术中的GroupBy和Sum?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我尝试将表中的几个字段分组,然后将这些组相加,但它们被重复计数。

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,它显示后面的SQL语句。在我看来是正确的,但是我不太熟悉SQL。问题是它输出为SUM_1的值被多次计数。如果我在"In Statement"中有[1]项,它将把这三个项相加。如果我有[1,2],它会把所有三个加起来,然后加倍,如果我有[1,2,3],它会把所有三个加起来,再加三倍。我不确定为什么它会计算多次。我如何修复此问题?

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!

推荐答案

这里有一些问题;您似乎没有查询正确的内容。按Expense.date分组时,选择费用对象是没有意义的。成本中心和费用之间需要某些联接条件,否则将复制行,每个成本中心都有计数,但两者之间没有关系。

您的查询应该如下所示:

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()

生成此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

这里有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))

这篇关于SQL炼金术中的GroupBy和Sum?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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