将sqlalchemy 查询结果转化成json格式

ALiDan 2019-11-05

models.py

class CompanyGoodsModel(Base):
    id=Column(Integer, primary_key=True)
    company_id = Column(Integer)
    。。。
    #我们在model中定义一个to_json方法,来获取模型的数据
    def to_json(self):
        dict = self.__dict__
        if "_sa_instance_state" in dict:
            del dict["_sa_instance_state"]
            return dict
    def get_info_by_id(company_id, category_id):
        if company_id:
            data=CompanyGoodsModel.query.filter(CompanyGoodsModel.company_id == company_id).all()
            return data
        elif category_id:
            data = CompanyGoodsModel.query.filter(CompanyGoodsModel.category_id == category_id).all()
            return data

然后在视图层中

@api.route(‘/company_goods‘, methods=[‘GET‘, ‘POST‘])
def company_goods():
    if request.method == ‘GET‘:
        company_id = request.args.get(‘company_id‘, ‘‘)
        category_id = request.args.get(‘category_id‘, ‘‘)
        data=CompanyGoodsModel.get_info_by_id(company_id, category_id)
        result=[]
        for d in data:
            d.goods_info=eval(d[‘goods_info‘])
            result.append(d.to_json())
        if not result:
            return json_response(‘0001‘, {}, ‘数据获取失败 ‘)
        return json_response(‘0000‘, result, ‘数据获取成功‘)

相关推荐

84593973 / 0评论 2020-06-14