设置作为参数传递的TypeScript对象的默认值

Question

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'});

我曾经想象过这样的事情可能有用:

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

显然,如果这些是明确的论据,你可以这样做:

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

sayName('bob');

在coffeescript中,您可以访问条件存在运算符,因此可以执行以下操作:

param.lastName ?= 'smith'

哪个编译成javascript:

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

Answer 1

实际上,现在似乎有一种简单的方法.以下代码适用于TypeScript 1.5:

function sayName({first,last='Smith'}: {first: string; last?: string}){
    var name = first + " " + last;
    alert(name);
}

sayName({firstName: 'Bob'});

诀窍是首先在括号中放入要从参数对象中选择的键,key=value对于任何默认值.使用:和类型声明跟随它.

这与您尝试执行的操作略有不同,因为params您不是拥有完整的对象,而是使用解除引用的变量.

如果要使其可选地将任何内容传递给函数,请?为类型中的所有键添加一个,并={}在类型声明后添加默认值:

function sayName({first='Bob',last='Smith'}: {first?: string; last?: string}={}){
    var name = first + " " + last;
    alert(name);
}

sayName();

Answer 2

Typescript现在支持默认参数:

https://www.typescriptlang.org/docs/handbook/functions.html

此外,添加默认值允许您省略类型声明,因为它可以从默认值推断:

function sayName(firstName: string, lastName = "Smith") {
  const name = firstName + ' ' + lastName;
  alert(name);
}

sayName('Bob');

Answer 3

不,TypeScript没有一种自然的方法来为一个对象的属性设置默认值,就像一个对象具有默认值而另一个没有默认值.您可以定义更丰富的结构:

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

    }
}

并使用它代替内联类型定义.

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

不幸的是,你不能做这样的事情:

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

} 

由于它只会设置默认值,如果name是undefined.

Answer 4

对象解构参数对象是上面的许多答案的目标,而现在,Typescript已经有了方法,使其更容易阅读和直观理解.

解构基础:通过解构对象,您可以按键名称从对象中选择属性.您可以定义您喜欢的少数或多个属性,默认值由基本语法设置let {key = default} = object.

let {firstName, lastName = 'Smith'} = myParamsObject;

//Compiles to:
var firstName = myParamsObject.firstName, 
_a = myParamsObject.lastName, 
lastName = _a === void 0 ? 'Smith' : _a;

为参数对象编写接口,类型或类可以提高易读性.

type FullName = {
  firstName: string;
   
  /** @default 'Smith' */
  lastName ? : string;
}

function sayName(params: FullName) {

  // Set defaults for parameter object
  var { firstName, lastName = 'Smith'} = params;

  // Do Stuff
  var name = firstName + " " + lastName;
  alert(name);
}

// Use it
sayName({
  firstName: 'Bob'
});

Answer 5

这可能是一个不需要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'}))

Answer 6

这里有一些可以尝试的方法，使用接口和使用默认值进行解构。请注意，“姓氏”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" })