如何在TypeScript中重载一个带有命名参数的函数?

Question

我有一个函数，我希望能够传递正常参数和关键字(命名)参数。然而，每当我尝试重载时，我都会得到一个错误。

我试过这样重载函数:

function myFunc({name, surname, age}: myFuncParams): string;
function myFunc(name: string, surname?: string, age?: number): string;
function myFunc(nameOrObj: string | myFuncParams, surname?: string, age?: number): string {
// do something and return a string
return ""
};

地点:

interface myFuncParams {
name: string,
surname?: string,
age?: number
}

当然，这是一个更复杂的情况的简化;但这里的想法是能够传递姓名和年龄，例如;无需移动姓氏或做以下操作:

myFunc("Foo", null, 111);

myFunc({name:"Foo", age:111});

我得到以下错误:

function myFunc({name, surname?, age?}: myFuncParams): string;
//        ^: This overload signature is not compatible with its implementation signature. (2394)
function myFunc(name: string, surname?: string, age?: number): string;
function myFunc(nameOrObj: string | myFuncParams, surname?: string, age?: number): string {
// do something and return a string
return ""
};

Typescript playground link:

编辑:正如@jonrsharpe所说:

function myFunc({name, surname, age}: myFuncParams): string;
function myFunc(name: string, surname?: string, age?: number): string;
function myFunc(nameOrObj: string | myFuncParams,surname?: string,age?: number): string {
return ""
};

忘记在最后声明中将姓氏和年龄标记为可选。

Answer 1

试试这个

function myFunc(opts: myFuncParams): string;
function myFunc(name: string, surname?: string, age?: number): string;
function myFunc(nameOrObj: string | myFuncParams, surname?: string, age?: number): string {
// do something and return a string
return ""
};