Question

当我尝试向构造函数分配静态属性时，打字稿出现错误：Property 'wheels' does not exist on type '() => void'.如何告诉打字稿我的Car对象可以有wheels财产？

function Car() {
    // do something
}

Car.wheels = 4 // throws: Property 'wheels' does not exist on type '() => void'.

const audi = new Car()

以上代码段可以在https://www.typescriptlang.org/play/index.html上进行测试

Answer 1

function Car() {
    // do something
}

namespace Car { 
    export let wheels = 4
}

const audi = new Car()

请参见handbook。

Answer 2

使用一些杂技类型，重命名和多个变量，您可以使用type assertion来实现所需的功能：

function Car_() {
    // do something
};

type TCar = typeof Car_ & { wheels: number };
const Car = (Car_ as TCar);

Car.wheels = 4; // throws: Property 'wheels' does not exist on type '() => void'.

const audi = new Car()

@MattMcCutchen建议的解决方案看起来更合法。

如何在构造函数上为静态属性定义类型

2 个答案: