我正在使用 Express 和 Seqeulize 运行 NodeJS,并且有一个文件 controllers/rooms.js
从 models/room.js
导入 Room。
import Room from '../models'
export function list(req, res) {
return Room
.findAll()
.then((rooms) => res.status(200).send(rooms))
.catch((error) => res.status(400).send(error))
}
波纹管是models/room.js
(同一目录下还有sequelize-cli生成的index.js
文件)
'use strict'
export default (sequelize, DataTypes) => {
const Room = sequelize.define('Room', {
name: DataTypes.STRING
})
return Room
}
我有一条路线 app.get('/rooms', list)
,但是当我访问此路线时,出现此错误:
TypeError: _models2.default.findAll is not a function
at list (/Users/matis/Documents/apps/node-docker-test/app/database/controllers/rooms.js:21:10)
at Layer.handle [as handle_request] (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/layer.js:95:5)
at next (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/route.js:137:13)
at Route.dispatch (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/route.js:112:3)
at Layer.handle [as handle_request] (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/layer.js:95:5)
at /Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:281:22
at Function.process_params (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:335:12)
at next (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:275:10)
at expressInit (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/middleware/init.js:40:5)
at Layer.handle [as handle_request] (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/layer.js:95:5)
at trim_prefix (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:317:13)
at /Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:284:7
at Function.process_params (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:335:12)
at next (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:275:10)
at query (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/middleware/query.js:45:5)
at Layer.handle [as handle_request] (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/layer.js:95:5)
at trim_prefix (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:317:13)
at /Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:284:7
at Function.process_params (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:335:12)
at next (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:275:10)
at Function.handle (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/router/index.js:174:3)
at Function.handle (/Users/matis/Documents/apps/node-docker-test/node_modules/express/lib/application.js:174:10)
我确定我的导入/导出搞砸了,但我不知道是怎么回事。
models/index.js
文件如下
'use strict'
import { readdirSync } from 'fs'
import { basename as _basename, join } from 'path'
import Sequelize from 'sequelize'
const basename = _basename(__filename)
const env = process.env.NODE_ENV || 'development'
const config = require(__dirname + '/../config/config.json')[env]
const db = {}
let sequelize
if (config.use_env_variable) {
sequelize = new Sequelize(process.env[config.use_env_variable], config)
} else {
sequelize = new Sequelize(config.database, config.username, config.password, config)
}
readdirSync(__dirname)
.filter(file => {
return (file.indexOf('.') !== 0) && (file !== basename) && (file.slice(-3) === '.js')
})
.forEach(file => {
const model = sequelize['import'](join(__dirname, file))
db[model.name] = model
})
Object.keys(db).forEach(modelName => {
if (db[modelName].associate) {
db[modelName].associate(db)
}
})
db.sequelize = sequelize
db.Sequelize = Sequelize
export default db
当我这样调用它时它就起作用了:
返回 Room.Room.findAll()...
因此我可以将导入重命名为:
从“../models”导入模型
并这样调用它:
返回 models.Room.findAll()...
但是为什么我不能直接调用它return Room.findAll()...
,导入应该如何制定??
最佳答案
假设您已设置 babel 来使用 ES6 导入,您可以尝试使用此方法在 ES6 中导出 Sequelize 模型。
//数据库配置文件
import Sequelize from 'sequelize';
export const sequelize = new Sequelize(
config.database.name,
config.database.user,
config.database.password,
{
host: config.database.host,
dialect: config.database.dialect,
pool: config.database.pool,
operatorsAliases: false
}
);
//模型
import Sequelize from 'sequelize';
import { sequelize } from '../database/db';
const User = sequelize.define(
'table_name',
{
id: {
type: Sequelize.INTEGER,
primaryKey: true,
autoIncrement: true
},
name: {
type: Sequelize.STRING,
allowNull: false
},
email: {
type: Sequelize.STRING,
allowNull: false
},
password: {
type: Sequelize.STRING,
allowNull: false
}
},
{ freezeTableName: true }
);
export default User;
关于node.js - NodeJS 无法导入 Sequelize.js 模型(ES6),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52987837/