在Typescript中,有没有一种方法可以使字符串数组的行为类似于“或拆分类型”?

我最近不得不这么做,我们有两个选择。

2. 一个数组和一个接口,所以每当您向标记中添加一个新项时,您都会同时添加到数组和对象中

这似乎是个人喜好的问题

我会用一个物体 Tags 强制执行类型安全

class Tags {
    public one: string;
    public two: string;
    public three: string;
    constructor({ one, two, three }: Tags) {
        this.one = one;
        this.two = two;
        this.three = three;
    }
}

interface Shape {
    name: string;
    tags: Tags;
}

let a = new Tags({}); // <-- This will fail because is not type Tag
a = new Tags({ one: '1' }); // <-- This will fail because it hasn't type two or type three
a = new Tags({ one: '1', two: '2', three: '3' }); // <-- This will pass

我们做的第二个选择如下:

const TagsArray = ["big", "small", "sharp", "dull"]
interface Tags {
  big: string;
  small: string;
  sharp: string;
  dull: string;
}

无论我们在哪里使用它们或传递标记,我们都会将其作为对象传递:

interface Shape {
  name: string;
  tags: Tags;
}

depends on your application ,对于我们来说,我们需要将一个数组传递给一些服务,并对以数组值作为属性的对象进行类型检查,因此我们选择了2,因为这样可以防止通过将数组转换为键 Object.keys(Tags) 无论服务返回什么,都使用 Shape

1. 这个 Tags 类型,它是标记可以是的每个字符串的联合类型
2. 这个 tagList {[key in Tags]: null} . 此对象类型要求每个标记都表示为一个属性(不再)。

export type Tags = "featured" | "design" | "fabrication" | "development" | "audio" 
| "imagery" | "social" | "leadership" | "writing" | "3d" | "interactive" | "work";

// the following typedef will assert that every tag is added as a prop
const tagsObject: {[key in Tags]: null} = {
    featured: null,
    design: null,
    fabrication: null,
    development: null,
    audio: null,
    imagery: null,
    social: null,
    leadership: null,
    writing: null,
    "3d": null,
    interactive: null,
    work: null,
}
// because the tagsObject must have all tags as props to pass type assertion
// tagList will always contain every tag
const tagList: Tags[] = Array.from(Object.keys(tagsObject)) as any;
export { tagList };