子查询与SQLAlchemy的

计数鉴于这些SQLAlchemy的模型定义：子查询与SQLAlchemy的

class Store(db.Model): 
    __tablename__ = 'store' 

    id = Column(Integer, primary_key=True) 
    name = Column(String, nullable=False) 


class CustomerAccount(db.Model, AccountMixin): 
    __tablename__ = 'customer_account' 

    id = Column(Integer, primary_key=True) 
    plan_id = Column(Integer, ForeignKey('plan.id'), index=True, nullable=False) 

    store = relationship('Store', backref='account', uselist=False) 
    plan = relationship('Plan', backref='accounts', uselist=False) 


class Plan(db.Model): 
    __tablename__ = 'plan' 

    id = Column(Integer, primary_key=True) 
    store_id = Column(Integer, ForeignKey('store.id'), index=True) 
    name = Column(String, nullable=False) 
    subscription_amount = Column(Numeric, nullable=False) 
    num_of_payments = Column(Integer, nullable=False) 
    store = relationship('Store', backref='plans')

如何编写一个查询的计划，让订阅收入的明细？我想找回给定商店的计划清单，并为每个计划计算该计划的总收入，计算方法是将Plan.subscription_amount * Plan.num_of_payments *订阅该计划的客户数量相乘计算得出

目前，我试图与此查询和子查询：

store = db.session.query(Store).get(1) 

subscriber_counts = db.session.query(func.count(CustomerAccount.id)).as_scalar() 

q = db.session.query(CustomerAccount.plan_id, func.sum(subscriber_counts * Plan.subscription_amount * Plan.num_of_payments))\ 
    .outerjoin(Plan)\ 
    .group_by(CustomerAccount.plan_id)

问题是子查询不能在当前计划的ID过滤。

我也试图与该另一种方法（无子查询）：

q = db.session.query(CustomerAccount.plan_id, func.count(CustomerAccount.plan_id) * Plan.subscription_amount * Plan.num_of_payments)\ 
    .outerjoin(Plan)\ 
    .group_by(CustomerAccount.plan_id, Plan.subscription_amount, Plan.num_of_payments)

虽然结果似乎很好，我不知道如何回到计划名称或其他计划列，如我d需要通过（并改变结果）将它们添加到组中。

理想情况下，如果一个计划没有任何订阅者，我希望它返回总数为零。

谢谢！

来源

2016-09-30 fabio.sussetto

感谢Alex格隆霍姆在#sqlalchemy我结束了这个工作的解决方案：

from sqlalchemy.sql.expression import label 
from sqlalchemy.sql.functions import coalesce 

from instalment.models import db 
from sqlalchemy import func, desc 


def projected_total_money_volume_breakdown(store): 
    subscriber_counts = db.session.query(CustomerAccount.plan_id, func.count(CustomerAccount.id).label('count'))\ 
     .group_by(CustomerAccount.plan_id)\ 
     .subquery() 

    total_amount_exp = coalesce(subscriber_counts.c.count, 0) * Plan.subscription_amount * Plan.num_of_payments 

    return db.session.query(Plan, label('total_amount', total_amount_exp))\ 
     .outerjoin(subscriber_counts, subscriber_counts.c.plan_id == Plan.id)\ 
     .filter(Plan.store == store)\ 
     .order_by(desc('total_amount'))\ 
     .all()

来源

2016-09-30 16:06:21

子查询与SQLAlchemy的

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