可以通过将成员添加到现有对象

扩大类型(读取:扩展/专门化接口) 将原始对象转换为扩展类型 将成员添加到对象中

interface IEnhancedPromise<T> extends Promise<T> {
    sayHello(): void;
}

const p = Promise.resolve("Peter");

const enhancedPromise = p as IEnhancedPromise<string>;

enhancedPromise.sayHello = () => enhancedPromise.then(value => console.info("Hello " + value));

// eventually prints "Hello Peter"
enhancedPromise.sayHello();

我倾向于把任何放在另一边，即var foo:IFoo = <任何>{};所以这样的东西仍然是类型安全的:

interface IFoo{
    bar:string;
    baz:string;
    boo:string;     
}

// How I tend to intialize 
var foo:IFoo = <any>{};

foo.bar = "asdf";
foo.baz = "boo";
foo.boo = "boo";

// the following is an error, 
// so you haven't lost type safety
foo.bar = 123; 

或者你可以将这些属性标记为可选:

interface IFoo{
    bar?:string;
    baz?:string;
    boo?:string;    
}

// Now your simple initialization works
var foo:IFoo = {};

在网上试试

只需这样做，您就可以添加或使用任何属性。(我使用的typescript版本为“typescript”:“~4.5.5”)

let contextItem = {} as any;

现在，您可以添加任何属性并在任何地方使用它。就像

contextItem.studentName = "kushal";

之后你可以这样使用它:

console.log(contextItem.studentName);

可以通过将成员添加到现有对象

扩大类型(读取:扩展/专门化接口) 将原始对象转换为扩展类型 将成员添加到对象中

interface IEnhancedPromise<T> extends Promise<T> {
    sayHello(): void;
}

const p = Promise.resolve("Peter");

const enhancedPromise = p as IEnhancedPromise<string>;

enhancedPromise.sayHello = () => enhancedPromise.then(value => console.info("Hello " + value));

// eventually prints "Hello Peter"
enhancedPromise.sayHello();

或者一气呵成:

  var obj:any = {}
  obj.prop = 5;

最好的做法是使用安全输入，我建议你:

interface customObject extends MyObject {
   newProp: string;
   newProp2: number;
}