【问题标题】:Get "insert_id" for one to one relationship in Flask, SqlAlchemy在 Flask、SqlAlchemy 中为一对一关系获取“insert_id”
【发布时间】:2015-06-11 07:56:07
【问题描述】:

我正在尝试通过 sqlalchemy 插入新行。父表 (Milestone) 有一个名为 Funding 的子表。两个表都通过一个名为milestone_id 的列共享关系。这是一对一的关系。

我已经查过了,但我不知道在 Funding 表中插入新行时如何引用里程碑 ID。父 ID 是自动增量的。我正在使用 Flask 和 SqlAlchemy。

型号:

 class Milestone(db.Model):
    __tablename__ = "**************"

   milestone_id = db.Column(db.Integer, primary_key=True)
   company_id = db.Column(db.Integer, db.ForeignKey('stlines_startups.company_id'))
   milestone_date = db.Column(db.Integer)
   snapshots = db.relationship('Snapshot', uselist=False, primaryjoin='Milestone.milestone_id==Snapshot.milestone_id', backref='milestone')
   fundraising = db.relationship('Funding', uselist=False, primaryjoin='Milestone.milestone_id==Funding.milestone_id', backref='milestone')

   def __init__(self, milestone_id, company_id, milestone_date, snapshots = [], fundraising = []):
    self.milestone_id = milestone_id
    self.company_id = company_id
    self.milestone_date = milestone_date
    self.snapshots = snapshots
    self.fundraising = fundraising

class Funding(db.Model):
    __tablename__ = "**************************"

   funding_id = db.Column(db.Integer, primary_key=True)
   funding_type = db.Column(db.Text)
   funding_message = db.Column(db.Text)
   funding_amount = db.Column(db.Integer)
   milestone_source = db.Column(db.Text)
   company_id = db.Column(db.Integer, db.ForeignKey('stlines_milestones.company_id'))
   milestone_id = db.Column(db.Integer, db.ForeignKey('stlines_milestones.milestone_id'))
   user_id = db.Column(db.Integer)
   funding_timestamp = db.Column(db.Integer)

def __init__(self, funding_id, funding_type, funding_message, funding_amount, milestone_source, milestone_id, company_id, user_id, funding_timestamp):
    self.funding_id = funding_id
    self.funding_type = funding_type
    self.funding_message = funding_message
    self.funding_amount = funding_amount
    self.milestone_source = milestone_source
    self.milestone_id = milestone_id
    self.company_id = company_id
    self.user_id = user_id
    self.funding_timestamp = funding_timestamp

炼金查询:

@app.route('/_add_funding')
def add_funding():
    funding_type = request.args.get('funding_stage', '', type=str)
    funding_message = request.args.get('funding_message', '', type=str)
    funding_amount = request.args.get('funding_amount', 0, type=int)
    milestone_source = request.args.get('milestone_source', '', type=str)
    milestone_date = request.args.get('milestone_date', '', type=str)
    company_id = request.args.get('company_id', '', type=int)

    milestone_date_final = datetime.datetime.strptime(milestone_date, '%B %d, %Y')

    ''' In this line, I try to reference milestone_id with new_milestone.milestone_id, but nothing shows up in the database '''
    new_funding = Funding('', funding_type=funding_type, funding_message=funding_message, funding_amount=funding_amount, milestone_source=milestone_source, company_id=company_id, milestone_id=new_milestone.milestone_id, user_id='1', funding_timestamp=milestone_date_final)
    new_milestone = Milestone('', company_id=company_id, milestone_date=milestone_date_final, snapshots=None, fundraising=new_funding)
    db.session.add(new_milestone)
    output = new_milestone.milestone_id
    db.session.commit()

    return jsonify(result=output)

在资金表中插入资金信息时,如何告诉 SqlAlchemy 使用里程碑表中自动生成的里程碑 ID?这些应该是两个单独的查询吗?

更新:

我接受了 ThiefMaster 关于使用刷新功能的建议,但我仍然收到错误消息: UnboundLocalError:赋值前引用了局部变量“new_milestone”

这是更新后的代码:

@app.route('/_add_funding')
def add_funding():
    funding_type = request.args.get('funding_stage', '', type=str)
    funding_message = request.args.get('funding_message', '', type=str)
    funding_amount = request.args.get('funding_amount', 0, type=int)
    milestone_source = request.args.get('milestone_source', '', type=str)
    milestone_date = request.args.get('milestone_date', '', type=str)
    company_id = request.args.get('company_id', '', type=int)

    milestone_date_final = datetime.datetime.strptime(milestone_date, '%B %d, %Y')
    ''' In this line, I try to reference milestone_id with new_milestone.milestone_id, but nothing shows up in the database '''
    new_funding = Funding('', funding_type=funding_type, funding_message=funding_message, funding_amount=funding_amount, milestone_source=milestone_source, company_id=company_id, milestone_id=new_milestone.milestone_id, user_id='1', funding_timestamp=milestone_date_final)
    new_milestone = Milestone('', company_id=company_id, milestone_date=milestone_date_final, snapshots=None, fundraising=new_funding)
    db.session.add(new_milestone)
    db.session.commit()
    db.session.flush()
    output = new_milestone.milestone_id

    return jsonify(result=output)

有什么想法吗?

【问题讨论】:

  • commit() 然后请求flush() 是没有意义的,因为commit() 会单独调用flush()。这个想法是,当您执行flush() 时,SQLAlchemy 将填充 autoincrement 列的值,因此您可以访问 autoincrement 的值,以便在添加新子表时可以从子表中填充 parent_id。注意:当您向 SQLAlchemy 会话添加记录时,它不会(还)将记录写入数据库。在您触发提交或刷新后,它会立即执行所有操作。这有助于提高性能。

标签: flask flask-sqlalchemy


【解决方案1】:

在您的db.session.add(..) 呼叫之后添加db.session.flush()。这将导致 INSERT 被发送到数据库,然后您就可以访问该 ID。

【讨论】:

    【解决方案2】:

    我找不到确切的解决方案。如果有人想知道,我最终通过直接执行 SQL 来解决它。这并不理想,但它现在可以完成工作。我最终一次插入一行,下面是代码:

    @app.route('/_add_funding')
    def add_funding():
        funding_type = request.args.get('funding_stage', '', type=str)
        funding_message = request.args.get('funding_message', '', type=str)
        funding_amount = request.args.get('funding_amount', 0, type=int)
        milestone_source = request.args.get('milestone_source', '', type=str)
        milestone_date = request.args.get('milestone_date', '', type=str)
        company_id = request.args.get('company_id', '', type=int)
    
        milestone_date_final =  time.mktime(datetime.datetime.strptime(milestone_date, '%B %d, %Y').timetuple())
    
        sql = "INSERT INTO ******** (`milestone_id`,`company_id`, `milestone_date`) VALUES ('','{}','{}')".format(company_id, milestone_date_final)
        result = db.engine.execute(sql)
    
        milestone_id = result.lastrowid 
    
        sql = "INSERT INTO ****** (`funding_id`,`funding_type`, `funding_message`, `funding_amount`, `milestone_source`, `company_id`, `milestone_id`, `user_id`, `funding_timestamp`) VALUES ('','{}','{}','{}','{}','{}','{}','1', '{}')".format(funding_type, funding_message, funding_amount, milestone_source, company_id, milestone_id, milestone_date_final)
        result = db.engine.execute(sql)
    
        output = result.lastrowid
        return jsonify(result=output)
    

    【讨论】:

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