javascript - 序列化 - 类型错误 : Cannot read property 'define' of undefined

Question

编辑:我的代码中存在一行，该行未在我的示例中添加，已添加并用注释标记。

我正在为一个新项目设置服务器，并自 v5 以来第一次尝试使用 Sequelize。运行我的服务器时，抛出此错误。

/Users/me/dev/projects/my-project/node_modules/sequelize/lib/model.js:730
    options = Utils.merge(_.cloneDeep(globalOptions.define), options);
                                                    ^

TypeError: Cannot read property 'define' of undefined
    at Function.init (/Users/me/dev/projects/my-project/node_modules/sequelize/lib/model.js:730:53)
    at Object.<anonymous> (/Users/me/dev/projects/my-project/src/server/db/models/user.js:16:6)
    at Module._compile (internal/modules/cjs/loader.js:736:30)
    at Object.Module._extensions..js (internal/modules/cjs/loader.js:747:10)
    ...

在调用我的模型定义初始化期间发生错误:

const sequelize = require("../_db");

...

console.log('this runs');

class User extends Sequelize.Model {} // edit: this line was here
User.init(
  {
    avatar: {
      default: {
        type: Sequelize.URL,
        value: defaultAvatar
      },
      type: Sequelize.STRING
    },
    email: {
      allowNull: false,
      type: Sequelize.STRING,
      unique: true
    },
    id: {
      defaultValue: Sequelize.UUIDV1,
      primaryKey: true,
      type: Sequelize.UUID
    },
    name: {
      type: Sequelize.STRING
    },
    password: {
      allowNull: false,
      get() {
        return () => this.getDataValue("password");
      },
      type: Sequelize.STRING
    },
    salt: {
      get() {
        return () => this.getDataValue("salt");
      },
      type: Sequelize.STRING
    }
  },
  { sequelize, modelName: "user" }
);

console.log('this doesn't);

这对我来说非常神秘，我不太确定出了什么问题。

Answer 1

您需要包含用户的类定义。 我在您提供的代码中看不到它。 这可以解释为什么错误说它来自“未定义”

即

const Model = Sequelize.Model;    
class User extends Model {}

你可以这样做:

class User extends Sequelize.Model {}

看这里: https://sequelize.org/master/manual/models-definition.html

此外，您似乎正在使用 Sequelize 的默认模型加载器，您需要在

中定义该模型

module.exports = (sequelize, DataTypes) => {
  const User = sequelize.define("User", {

  });

  return User
};