Arrays 类型脚本数组分配错误

我有这门课：

user.ts

const ADMIN = 1;
const AGENT = 2;
const ADJUSTER = 3;
const MANAGER = 4;
const MOTORCYCLIST = 5;

export class User {
  id: number;
  name: string;
  email: string;
  phone: string;
  status: string;
  role: number;

  isAdmin(): boolean {
    return this.role === ADMIN;
  }

  isAgent(): boolean {
    return this.role === AGENT;
  }

  isAdjuster(): boolean {
    return this.role === ADJUSTER;
  }

  isManager(): boolean {
    return this.role === MANAGER;
  }

  isMotorcyclist(): boolean {
    return this.role === MOTORCYCLIST;
  }
}

import { User } from '../models/user';

export const USERS: User[] = [{"id":301,"name":"Shirlee Zboncak","email":"kenyatta_jerde@erdman.ca","phone":"044 93 1342 0977","status":"inactive","role":3},{"id":302,"name":"Erich Auer","email":"grazyna.romaguera@bogisich.name","phone":"044 87 8444 6879","status":"inactive","role":2},{"id":303,"name":"Jerrold McDermott","email":"lindsey_brekke@beierdickinson.info","phone":"044 64 7490 1751","status":"inactive","role":3},{"id":304,"name":"Yael Hilpert","email":"stanton.abbott@pouros.co.uk","phone":"044 16 9190 7970","status":"inactive","role":4},{"id":305,"name":"Pat Hodkiewicz","email":"elin@sanford.biz","phone":"044 12 0242 7692","status":"inactive","role":3}]

出于测试原因，我使用了一个mock users数组（为了简单起见，我将显示一个只有5个用户的数组，但在我的代码中我有1000多个用户）：

模拟用户。ts

const ADMIN = 1;
const AGENT = 2;
const ADJUSTER = 3;
const MANAGER = 4;
const MOTORCYCLIST = 5;

export class User {
  id: number;
  name: string;
  email: string;
  phone: string;
  status: string;
  role: number;

  isAdmin(): boolean {
    return this.role === ADMIN;
  }

  isAgent(): boolean {
    return this.role === AGENT;
  }

  isAdjuster(): boolean {
    return this.role === ADJUSTER;
  }

  isManager(): boolean {
    return this.role === MANAGER;
  }

  isMotorcyclist(): boolean {
    return this.role === MOTORCYCLIST;
  }
}

import { User } from '../models/user';

export const USERS: User[] = [{"id":301,"name":"Shirlee Zboncak","email":"kenyatta_jerde@erdman.ca","phone":"044 93 1342 0977","status":"inactive","role":3},{"id":302,"name":"Erich Auer","email":"grazyna.romaguera@bogisich.name","phone":"044 87 8444 6879","status":"inactive","role":2},{"id":303,"name":"Jerrold McDermott","email":"lindsey_brekke@beierdickinson.info","phone":"044 64 7490 1751","status":"inactive","role":3},{"id":304,"name":"Yael Hilpert","email":"stanton.abbott@pouros.co.uk","phone":"044 16 9190 7970","status":"inactive","role":4},{"id":305,"name":"Pat Hodkiewicz","email":"elin@sanford.biz","phone":"044 12 0242 7692","status":"inactive","role":3}]

问题是，该代码在以下情况下失败：

在我的编辑器中，它显示了另一条消息：

Missing property 'isAdmin' on type '{ "id": number; "name": string; "email": string; "phone": string; "status": string; "role": numbe...'.

---

如何使我的用户类不尝试分配方法。我只想分配那些id、姓名、电子邮件、电话、状态和角色属性。我尝试使用构造函数，但错误相同。

您可以将构造函数添加到类中，然后只映射数组：

constructor(id,name,email,phone,status,role) {
    this.id = id;
    //...etc
}

然后

---

你说你试过使用构造函数，但也许你做得不对

这里有一个方法可以解决你的问题。首先，创建一个接口来统一构建

user

对象所需的用户数据

export interface UserData {
  id: number;
  name: string;
  email: string;
  phone: string;
  status: string;
  role: number;
}

然后在

User

类定义中，声明一个类型为

UserData

的字段，该字段将保存用户数据

export class User {
  data: UserData;

  constructor(userData: UserData) {
    this.data = userData;
  }

  isAdmin(): boolean {
    return this.data.role === ADMIN;
  }

  // + all other method declarations
}

如果不想将用户数据存储到类型为

UserData

的字段中，可以使用构造函数中的接口，如下所示：

export class User {
  id: number;
  name: string;
  email: string;
  phone: string;
  status: string;
  role: number;

  constructor(userData: UserData) {
    this.id = userData.id;
    this.name = userData.name;
    // etc.
  }

  isAdmin(): boolean {
    return this.role === ADMIN;
  }
}

然后，您可以使用

new

关键字和构造函数实例化

User

类中的对象，构造函数将

UserData

对象作为参数

export const USERS: User[] = [
  new User({
    id: 301,
    name: "Shirlee Zboncak",
    email: "kenyatta_jerde@erdman.ca",
    phone: "044 93 1342 0977",
    status: "inactive",
    role: 3
  })
];

---

现在您可以创建

User

对象，而无需指定方法定义。

“我曾尝试使用构造函数，但错误相同。”我对此表示怀疑；请告诉我们。@jcalz发生的事情是，我刚刚创建了构造函数，但没有做任何其他事情。这看起来是一种非常酷的方法。我会使用它，我会让你知道它是否适合我。如果你想保留现有的类声明字段结构，我会添加一个替代类定义。谢谢！我想知道怎么做。我试过你的方法，效果很好！