javascript - 继承不存在的 ES6 模型方法

Question

在尝试将 Sequelize JS v4 与 ES6 类一起使用时，我在执行实例方法时遇到了问题。出于某种原因，尽管在代码中定义了它们，但它们似乎并不存在。

这是一个例子-

模型文件

'use strict';
const Sequelize = require("sequelize");

class Model extends Sequelize.Model {

  static init(sequelize, DataTypes) {
    return super.init(
      {
        // properties
      },
      { sequelize }
    );
  }

  static associate(models) {
  }


  async modelMethod() {

  }

}

module.exports = Model;

模型启动

let modelClass = require('../models/' + modelFile);

let model = modelClass.init(sequelize, Sequelize);

然后模型作为 Controller 的属性在 Controller 文件中被调用

async controllerMethod(req, res) {
    let info = await this.model.modelMethod();
    res.send(info);
}

以及我收到的错误 -

TypeError: this.model.modelMethod is not a function at Controller.controllerMethod (/usr/app/controllers/controller.js:83:41) at Layer.handle [as handle_request] (/usr/app/node_modules/express/lib/router/layer.js:95:5) at next (/usr/app/node_modules/express/lib/router/route.js:137:13) at Route.dispatch (/usr/app/node_modules/express/lib/router/route.js:112:3) at Layer.handle [as handle_request] (/usr/app/node_modules/express/lib/router/layer.js:95:5) at /usr/app/node_modules/express/lib/router/index.js:281:22 at param (/usr/app/node_modules/express/lib/router/index.js:354:14) at param (/usr/app/node_modules/express/lib/router/index.js:365:14) at param (/usr/app/node_modules/express/lib/router/index.js:365:14) at Function.process_params (/usr/app/node_modules/express/lib/router/index.js:410:3) at next (/usr/app/node_modules/express/lib/router/index.js:275:10) at sequelize.models.Session.findOne.then (/usr/app/app.js:44:24) at tryCatcher (/usr/app/node_modules/bluebird/js/release/util.js:16:23) at Promise._settlePromiseFromHandler (/usr/app/node_modules/bluebird/js/release/promise.js:512:31) at Promise._settlePromise (/usr/app/node_modules/bluebird/js/release/promise.js:569:18) at Promise._settlePromise0 (/usr/app/node_modules/bluebird/js/release/promise.js:614:10) at Promise._settlePromises (/usr/app/node_modules/bluebird/js/release/promise.js:693:18) at Async._drainQueue (/usr/app/node_modules/bluebird/js/release/async.js:133:16) at Async._drainQueues (/usr/app/node_modules/bluebird/js/release/async.js:143:10) at Immediate.Async.drainQueues (/usr/app/node_modules/bluebird/js/release/async.js:17:14) at Immediate.args.(anonymous function) [as _onImmediate] (/usr/local/lib/node_modules/pm2/node_modules/event-loop-inspector/index.js:133:29) at runCallback (timers.js:810:20)

尝试输出类的方法得到这个 -

console.log(Object.getOwnPropertyNames(this.model));


[ 'length',
  'prototype',
  'init',
  'associate',
  'name',
  'sequelize',
  'options',
  'associations',
  'underscored',
  'tableName',
  '_schema',
  '_schemaDelimiter',
  'rawAttributes',
  'primaryKeys',
  '_timestampAttributes',
  '_readOnlyAttributes',
  '_hasReadOnlyAttributes',
  '_isReadOnlyAttribute',
  '_dataTypeChanges',
  '_dataTypeSanitizers',
  '_booleanAttributes',
  '_dateAttributes',
  '_hstoreAttributes',
  '_rangeAttributes',
  '_jsonAttributes',
  '_geometryAttributes',
  '_virtualAttributes',
  '_defaultValues',
  'fieldRawAttributesMap',
  'fieldAttributeMap',
  'uniqueKeys',
  '_hasBooleanAttributes',
  '_isBooleanAttribute',
  '_hasDateAttributes',
  '_isDateAttribute',
  '_hasHstoreAttributes',
  '_isHstoreAttribute',
  '_hasRangeAttributes',
  '_isRangeAttribute',
  '_hasJsonAttributes',
  '_isJsonAttribute',
  '_hasVirtualAttributes',
  '_isVirtualAttribute',
  '_hasGeometryAttributes',
  '_isGeometryAttribute',
  '_hasDefaultValues',
  'attributes',
  'tableAttributes',
  'primaryKeyAttributes',
  'primaryKeyAttribute',
  'primaryKeyField',
  '_hasPrimaryKeys',
  '_isPrimaryKey',
  'autoIncrementAttribute',
  '_scope',
  '_scopeNames' ]

Answer 1

阅读 Manish 的评论后，我决定重新尝试一下。

显然，实例方法的工作方式是在调用 .init 方法后初始化模型。

所以 -

let model = ModelClass.init(sequelize, Sequelize);
model = new model();

这将允许毫无问题地调用 model.modelMethod