Question

构建以下打字稿代码时，出现以下错误。

Property 'runStaticMethod' is a static member of type 'DemoClass'

打字稿代码：


export class Main {
    constructor(private demo: DemoClass) { }
    public run() {
        this.demo.runStaticMethod();
    }
}

export class DemoClass {
    public static runStaticMethod() {
        console.log('run...');
    }
}

new Main(DemoClass).run();

构建上面的console代码时，出现以下typescript错误。但是javascript代码正在按预期运行。

控制台错误：

Chitty:tsc NatarajanG$ tsc
src/index.ts:5:19 - error TS2576: Property 'runStaticMethod' is a static member of type 'DemoClass'

5         this.demo.runStaticMethod();
                    ~~~~~~~~~~~~~~~

Chitty:tsc NatarajanG$

Answer 1

由于它是静态属性，因此尽管JavaScript支持DemoClass.runStaticMethod()，您仍应按照TS要求的方式访问它：this.demo.runStaticMethod()。

https://www.typescriptlang.org/docs/handbook/classes.html#static-properties

每个实例都通过在类名前添加访问此值。与前置this类似。在实例访问之前，这里我们在Grid之前。在静态访问之前。

Answer 2

我的测试表明，为 interface 添加 DemoClass 并为 runStaticMethod 定义可以解决错误，例如...

export interface DemoClass {
    runStaticMethod(...args: any[]): any;
}

...为了方便复制/粘贴，这里是进一步自定义的起点...

export interface DemoClass {
    runStaticMethod(...args: any[]): any;
}


export class Main {
    constructor(private demo: DemoClass) { }
    public run() {
        this.demo.runStaticMethod();
    }
}


export class DemoClass {
    public static runStaticMethod() {
        console.log('run...');
    }
}


new Main(DemoClass).run();

属性“ runStaticMethod”是类型“ DemoClass”的静态成员

2 个答案: