Question

我正在使用TS来构建节点快速API。我区分model和viewmodel类。 Post模型类可以有多个视图模型表示。

但是如何从模型转换为viewmodel类？我可以创建一个创建新viewmodel对象的方法。但是有可能只是转换为另一种类型吗？

示例

Post.ts

export interface IPost {
    id: number;
    author: string;
    heading: string;
    body: string;
}

/**
 * Class to model a blog post
 */
export class Post implements IPost {

    public id: number;
    public author: string;
    public heading: string;
    public body: string;

    constructor(id: number, author: string, heading: string, body: string) {
        this.id = id;
        this.author = author;
        this.heading = heading;
        this.body = body;
    }
}

PostVM.ts

export interface IPostVM {
    id: number;
    author: string;
    text?: string;
}

/**
 * Class to model a blog post
 */
export class PostVM implements IPostVM {

    public id: number;
    public author: string;
    public text?: string;

    constructor(id: number, author: string, body: string) {
        this.id = id;
        this.author = author;
        this.text = body;
    }
}

App.ts

// This is NOT working:
const post: IPost = this.dao.getPostById(id);
const postVM: IPostVM = post as PostVM;

结果：

{
    "author": "Ole",
    "body": "Dette er en test på body tekst.",
    "heading": "Overskrift 1",
    "id": 1
}

应该是：

{
    "id": 1
    "author": "Ole",
    "text": "Overskrift 1",
}

Answer 1

你不能只是＆＃34;演员＆＃34;因为，评论说，除非你自己写，否则没有办法知道你希望如何转换课程。

正如你所说，你可以做的最好的事情就是为班级编写一个函数：

export class Post implements IPost {
    <...>

    toPostVM() {
        return new PostVM(this.id, this.author, this.body);
    }
}

然后在你的App.ts中你可以做到：

const postVM: IPostVM = post.toPostVM();

打字稿：转换为另一种类似的类型

1 个答案: