Flask sqlalchemy.exc.NoForeignKeysError:Can'；在'；服务'；和'；修理'；

我试图将类图映射到数据库表中。我有一个

repair

类继承了

服务

类。但是当我尝试这样做时，它说，

在“服务”和“修复”之间找不到任何外键关系

这是我的全部代码

from flask import Flask, request, jsonify
from flask_sqlalchemy import SQLAlchemy
from flask_marshmallow import Marshmallow
from sqlalchemy import ForeignKey

import os

app = Flask(__name__)
basedir = os.path.abspath(os.path.dirname(__file__))

app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///' + os.path.join(basedir, 'db.sqlite')
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False

database = SQLAlchemy(app)
ORM = Marshmallow(app)


# Models
# Service Model
class Service(database.Model):
    service_id = database.Column(database.Integer, primary_key=True)
    image_url = database.Column(database.String(255))
    jewellery_type = database.Column(database.String(50))

    def __init__(self, image_url, jewellery_type):
        self.image_url = image_url
        self.jewellery_type = jewellery_type


# Service Schema
class ServiceSchema(ORM.Schema):
    class Meta:
        fields = ('service_id', 'image_url', 'jewellery_type')


# Initialize Service Schema
service_schema = ServiceSchema()
Service_schema_all = ServiceSchema(many=True)


# repair Model
class Repair(Service):
    repair_id = database.Column(database.Integer, ForeignKey('service_id'), primary_key=True)
    damage_type = database.Column(database.String(50))
    repair_type = database.column(database.String(50))

    def _init__(self, repair_id, damage_type, repair_type):
        self.repair_id = repair_id
        self.damage_type = damage_type
        self.repair_type = repair_type


# repair Schema
class RepairSchema(ORM.Schema):
    class Meta:
        fields = ('repair_id', 'damage_type', 'repair_type')


# Initialize Repair Schema
repair_schema = RepairSchema()
repair_schema_all = RepairSchema(many=True)



# start the server
if __name__ == '__main__':
    app.run(debug=True)

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尝试更改此引用

ForeignKey('service_id') 

到

---

我做了这么多的研究，仍然不能解决这个问题。

ForeignKey('service.service_id').