function sayName(params: {firstName: string; lastName?: string}) {
    params.lastName = params.lastName || 'smith';  // <<-- any better alternative to this?
    var name = params.firstName + params.lastName
    alert(name);
}

sayName({firstName: 'bob'});

我曾想象过这样的东西可能有用:

function sayName(params: {firstName: string; lastName: string = 'smith'}) {

显然,如果这些是简单的参数,你可以这样做:

function sayName(firstName: string, lastName = 'smith') {
    var name = firstName + lastName;
    alert(name);
}

sayName('bob');

在coffeescript中,你可以访问条件存在操作符,所以可以这样做:

param.lastName ?= 'smith'

编译成javascript:

if (param.lastName == null) {
    param.lastName = 'smith';
}

当前回答

还有另一种不需要解构的方法,就是用这样的类型设置默认值:

function name(param1:type=defaultValue1, paramr2:type=defaultvalue2,...) {//}

一个例子是

function addTwoNumbers(first:number = 1, second:number = 3):number {return first+second}
console.log(addTwoNumbers())

其他回答

这里有一些东西可以尝试,使用默认值的接口和解构。请注意,“lastName”是可选的。

interface IName {
  firstName: string
  lastName?: string
}

function sayName(params: IName) {
  const { firstName, lastName = "Smith" } = params
  const fullName = `${firstName} ${lastName}`

  console.log("FullName-> ", fullName)
}

sayName({ firstName: "Bob" })

还有另一种不需要解构的方法,就是用这样的类型设置默认值:

function name(param1:type=defaultValue1, paramr2:type=defaultvalue2,...) {//}

一个例子是

function addTwoNumbers(first:number = 1, second:number = 3):number {return first+second}
console.log(addTwoNumbers())

不,TypeScript没有一种自然的方法来为对象的属性设置默认值,比如一个有默认值,另一个没有。你可以定义一个更丰富的结构:

class Name {
    constructor(public first : string, 
        public last: string = "Smith") {

    }
}

用它来代替内联类型定义。

function sayName(name: Name) {
    alert(name.first + " " + name.last);
}

不幸的是,你不能这样做:

function sayName(name : { first: string; last?:string } 
       /* and then assign a default object matching the signature */  
       = { first: null, last: 'Smith' }) {

} 

因为它只会在名称未定义时设置默认值。

这是一种不涉及长构造函数的好方法

class Person {
    firstName?: string = 'Bob';
    lastName?: string = 'Smith';

    // Pass in this class as the required params
    constructor(params: Person) {
        // object.assign will overwrite defaults if params exist
        Object.assign(this, params)
    }
}

// you can still use the typing 
function sayName(params: Person){ 
    let name = params.firstName + params.lastName
    alert(name)
}

// you do have to call new but for my use case this felt better
sayName(new Person({firstName: 'Gordon'}))
sayName(new Person({lastName: 'Thomas'}))

不需要解构,您可以创建一个默认的params并将其传入

interface Name {
   firstName: string;
   lastName: string;
}

export const defaultName extends Omit<Name, 'firstName'> {
    lastName: 'Smith'
}

sayName({ ...defaultName, firstName: 'Bob' })