我正在使用 sequelize 访问 postgres 数据库,我想查询一个城市,例如包含“Building”表,但我想将输出重命名为“buildings”并返回 http 响应,但出现此错误:


    City.findById(req.params.id,{
      include: [
        {
          model: Building, as: "buildings"
        }
      ]
    }).then(city =>{
      console.log(city.id);
         res.status(201).send(city);
    }) .catch(error => {
     console.log(error);
     res.status(400).send(error)
   });

城市模型
            const models = require('../models2');
            module.exports = (sequelize, DataTypes) => {
              const City = sequelize.define('city', {
              name: { type: DataTypes.STRING, allowNull: false },
                status: { type: DataTypes.INTEGER, allowNull: false },
                latitude: { type: DataTypes.BIGINT, allowNull: false },
                longitude: { type: DataTypes.BIGINT, allowNull: false },

              }, { freezeTableName: true});
              City.associate = function(models) {
                // associations can be defined here
                 City.hasMany(models.building,{as: 'building', foreignKey: 'cityId'})
              };
              return City;
            };

最佳答案

正如您在下面的代码中定义的别名是 building :

City.hasMany(models.building,{as: 'building', foreignKey: 'cityId'})

但是在查询中,您使用的是 buildings
include: [
  {
     model: Building, as: "buildings" // <---- HERE
  }
]

它应该是 building :
include: [
   {
         model: Building, as: "building" // <---- HERE
   }
]

关于node.js - sequelize 选择并包含另一个表别名,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/53117988/

10-14 03:45