tit*_*aka 2 python sqlalchemy flask flask-sqlalchemy
我需要使用flask-sqlalchemy将两个模型员工和技能与额外数据相关联。
例如,用户1可以做skill1用效率为100%和skill2用效率80%。
所以,我制作了 3 个模型:
class Employee(db.Model):
__tablename__ = 'employees'
id = db.Column(db.Integer, primary_key=True)
login = db.Column(db.String(64), index=True, unique=True, nullable=False)
skills = db.relationship('Skill', secondary='employee_skills',
backref=db.backref('employees', lazy='dynamic'))
def __repr__(self):
return '<Employee %r, Skills: %r>' % (self.login, self.skills)
class Skill(db.Model):
__tablename__ = 'skills'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(64), nullable=False, unique=True)
def __repr__(self):
return '<Skill %r>' % self.name
class EmployeeSkill(db.Model):
__tablename__ = 'employee_skills'
employee_id = db.Column(db.Integer, db.ForeignKey('employees.id'), primary_key=True)
skill_id = db.Column(db.Integer, db.ForeignKey('skills.id'), primary_key=True)
efficiency = db.Column(db.Float)
skills_efficiency = db.relationship('Skill', backref=db.backref('employee_efficiency', lazy='dynamic'))
def __repr__(self):
return '<Engineer %r, skill %r, efficiency %r>' % (
self.employee_id, self.skill_id, self.efficiency)
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之后,我尝试访问此关联:
>>> from app import db, models
>>> emp_=models.Employee.query.all()
>>> emp_[0]
<Employee u'user1', Skills: [<Skill u'skill1'>, <Skill u'skill2'>]>
>>> emp_[0].skills
[<Skill u'skill1'>, <Skill u'skill2'>]
>>> emp_[0].skills[0]
<Skill u'skill1'>
>>> emp_[0].skills[0].employee_efficiency
<sqlalchemy.orm.dynamic.AppenderBaseQuery object at 0x0000000003A68FD0>
>>> emp_[0].skills[0].employee_efficiency[0]
<Engineer 1, skill 1, efficiency u'1'>
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我想要的是将Employee实例表示为
<Employee u'user1', Skills: <u'skill1', efficiency: '1'>, <u'skill2', efficiency: '0.8'>>
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但我不明白:我是否以错误的方式使用 db.relationship?还是我需要完善我的repr方法?
提前致谢