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[英]How to overload function in TypeScript with optional parameters?
[英]How do I overload a function with named parameters in TypeScript?
我有一个 function,我希望能够传递正常的 arguments 和关键字(命名)arguments。但是,每当我尝试进行重载时,我都会收到错误消息。
我尝试以这种方式重载 function:
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 ""
};
在哪里:
interface myFuncParams {
name: string,
surname?: string,
age?: number
}
当然,这是对更复杂的情况的简化; 但是这里的想法是能够传递姓名和年龄,例如; 无需移动姓氏或执行以下操作:
myFunc("Foo", null, 111);
只是在做:
myFunc({name:"Foo", age:111});
我收到以下错误:
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 游乐场链接:
编辑:正如@jonrsharpe 指出的那样:
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 ""
};
忘记在最终声明中将姓氏和年龄标记为可选。
试试这个方法
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 ""
};
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