Question

我是烧瓶和炼金术的新手，并决定通过使用烧瓶进行大型项目来学习。当我开始在类之间创建关系时。每个类都在单独的文件和文件夹中。我想要实现的逻辑是每个controlaccount只能分配一个项目，而项目可以分配给多个控制，即一对一的关系。

我一直收到以下错误消息 sqlalchemy.exc.InternalError：（pymysql.err.InternalError）（1054，“字段列表'中的未知列'project_id'”）[SQL：'INSERT INTO controlaccounts（代码，名称，预算，PMB_start，{ {1}}，PMB_finish，PMU_start，parent_id，project_id）值（％s，％s，％s，％s，％s，％s，％s，％s，％s） '] [参数：（'Root'，'Root'，1.0，datetime.date（2017,12,1），datetime.date（2017,12,31），datetime.date（2018,12,1），datetime .date（2019,2,27），1,1）]

我在controlaccounts / models.py中的代码

PMU_finish

项目/模型文件代码是

class ControlAccount(db.Model):
    __tablename__='controlaccounts'
    id = db.Column(db.Integer, primary_key =True)
    code = db.Column(db.String(80))
    name = db.Column(db.String(80))
    budget = db.Column(db.Float)
    PMB_start = db.Column(db.Date)
    PMB_finish = db.Column(db.Date)
    PMU_start = db.Column(db.Date)
    PMU_finish = db.Column(db.Date)
    parent_id = db.Column(db.Integer)
    project_id = db.Column(db.Integer, db.ForeignKey('projects.id'))


    def __init__(self, code, name, budget, PMB_start, PMB_finish, PMU_start, PMU_finish, parent_id, project):
        self.code = code
        self.name = name
        self.parent_id = parent_id
        #self.accounttype_id = accounttype.id
        self.budget = budget
        self.PMB_start = PMB_start
        self.PMB_finish = PMB_finish
        self.PMU_start = PMU_start
        self.PMU_finish = PMU_finish
        self.project_id = project.id


    def __repr__(self):
        return self.name

controlaccount代码的表单是

class Project(db.Model):


__tablename__ ="projects"
    id = db.Column(db.Integer, primary_key=True)
    code = db.Column(db.String(80))
    name = db.Column(db.String(80))
    owner = db.Column(db.Integer, db.ForeignKey('users.id'))
    description = db.Column(db.Text)
    start = db.Column(db.DateTime)
    finish = db.Column(db.DateTime)
    status = db.Column(db.Boolean)
    project = db.relationship('ControlAccount', backref='projects', lazy='dynamic',primaryjoin="Project.id == ControlAccount.project_id")

    def __init__(self, code, name, description, owner, start, finish, status):
        self.code = code
        self.name = name
        self.owner = owner
        self.description = description
        self.start = start
        self.finish = finish
        self.status = status

，最后是观看代码

class ControlAccountForm(Form):

    def get_projects():
        return Project.query.all()

    code = StringField('Code', [validators.Required()])
    name = StringField('Type Name', [validators.Required()])
    #accounttype = QuerySelectField('Account Type', query_factory= get_account_types)
    #curve_id = QuerySelectField('Spread Profile', query_factory= get_spread_profile)
    budget = FloatField('Budget', [validators.Required()])
    PMB_start = DateField('Planned Start', [validators.Required()])
    PMB_finish = DateField('Planned Finish', [validators.Required()])
    PMU_start = DateField('Anticipated Start')
    PMU_finish = DateField('Anticipated Finish')
    #parent_id = QuerySelectField('Parent Accounts', query_factory= get_control_accounts)
    parent_id = IntegerField('Parent ID')
    project_id = QuerySelectField('Project', query_factory= get_projects)

Answer 1

由于错误显示数据库中缺少project_id字段。如果已经创建了数据库查看器，请查看它。否则，删除数据库并使用db.create_all()重新创建它。

此外，project类中的Project字段也应删除。这没有意义。如果要获得相关控制帐户的backref，只需在ControlAccount类中添加backref。有关详细信息，请参阅http://docs.sqlalchemy.org/en/latest/orm/basic_relationships.html#one-to-one。

class ControlAccount(db.Model):
    __tablename__='controlaccounts'
    # ...
    project_id = db.Column(db.Integer, db.ForeignKey('projects.id'))
    project = relationship("Project", backref="control_account")

Flask-Alchemy关系错误

1 个答案: