Question

我正在使用Sequelize，并且有以下代码：

迁移文件：

'use strict'

module.exports = {
  up: (queryInterface, Sequelize) => {
    return queryInterface.createTable('equipments', {
      id: {
        type: Sequelize.INTEGER,
        primaryKey: true,
        autoIncrement: true
      },
      hostname: {
        type: Sequelize.STRING,
        allowNull: false,
        unique: true
      },
      vendor: {
        type: Sequelize.STRING,
        allowNull: false
      },
      model: {
        type: Sequelize.STRING,
        allowNull: false
      },
      equipment_class_id: {
        type: Sequelize.INTEGER,
        allowNUll: false,
        references: { model: 'equipment_classes' }
      },
      created_at: {
        type: Sequelize.DATE,
        allowNull: false
      },
      updated_at: {
        type: Sequelize.DATE,
        allowNull: false
      }
    })
  },

  down: (queryInterface, Sequelize) => {
    return queryInterface.dropTable('equipments')
  }
}

模型设备：

const { Model, DataTypes } = require('sequelize')

class Equipment extends Model {
  static init (sequelize) {
    super.init({
      hostname: DataTypes.STRING,
      vendor: DataTypes.STRING,
      model: DataTypes.STRING,
      equipmentClassId: {
        type: DataTypes.INTEGER,
        field: 'equipment_class_id'
      }
    }, {
      sequelize
    })
  }
}

module.exports = Equipment

在数据库中，该表被创建为设备名称。

CREATE TABLE public.equipments
(
    id integer NOT NULL DEFAULT nextval('equipments_id_seq'::regclass),
    hostname character varying(255) COLLATE pg_catalog."default" NOT NULL,
    vendor character varying(255) COLLATE pg_catalog."default" NOT NULL,
    model character varying(255) COLLATE pg_catalog."default" NOT NULL,
    equipment_class_id integer,
    created_at timestamp with time zone NOT NULL,
    updated_at timestamp with time zone NOT NULL,
    CONSTRAINT equipments_pkey PRIMARY KEY (id),
    CONSTRAINT equipments_hostname_key UNIQUE (hostname)
,
    CONSTRAINT equipments_equipment_class_id_fkey FOREIGN KEY (equipment_class_id)
        REFERENCES public.equipment_classes (id) MATCH SIMPLE
        ON UPDATE NO ACTION
        ON DELETE NO ACTION
)

当我运行此代码时：

async index (req, res) {
    const equipments = await Equipment.findAll()

    return res.json(equipments)
  }

我收到错误：

Executing (default): SELECT "id", "hostname", "vendor", "model", "equipment_class_id" AS "equipmentClassId", "created_at" AS "createdAt", "updated_at" AS "updatedAt" FROM "equipment" AS "Equipment";
(node:17032) UnhandledPromiseRejectionWarning: SequelizeDatabaseError: relation "equipment" does not exist

我的数据库：Postgresql 12

我还有其他类似创建的表，但是我没有这个问题，我在做什么错了？

Paulo

Answer 1

您的问题正在命名。 Sequelize使用模型的名称自动重写到数据库表中。

在迁移中，您具有：

return queryInterface.createTable('equipments'...

但是您的模型具有：

class Equipment extends Model {

命名约定将模型中的名称复数以在数据库中查找。因此，您应该在迁移文件中添加

：

return queryInterface.createTable('Equipment'...

设备是单数设备的复数形式。

您还可以使用sequelize-cli简化此过程。

使用sequelize-cli，您不必担心命名约定。只需运行（用您的属性列表替换属性列表）：

sequelize model:generate --name Equipment --attributes fun:string,done:integer

它应该为您处理复数形式，生成一个迁移和模型文件。

SequelizeDatabaseError：关系“设备”不存在

1 个答案: