Javascript es6使用typescript在函数中分解参数

我仍然希望使用解构，因为它可以帮助调用函数。但我在函数params中遇到了一个问题：

const t = ({id, name}: {number, string}) => `id: ${id} name:${name}`

另一个失败的方法是，我试图避免将所有类型的参数放在一行中，但使用了一个接口：

interface tParams {
    id: number,
    name: string
}
const t = ({id, name} as tParams) => `id: ${id} name:${name}`

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您可以这样做：

const t = ({id, name}: {id: number, name: string}) => `id: ${id} name:${name}`

interface tParams {
    id: number,
    name: string
}
const t = ({id, name}: tParams) => `id: ${id} name:${name}`

或者这样：

const t = ({id, name}: {id: number, name: string}) => `id: ${id} name:${name}`

interface tParams {
    id: number,
    name: string
}
const t = ({id, name}: tParams) => `id: ${id} name:${name}`

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Destructure适用于任何对象类型。所以，如果您的对象定义正确，解构将起作用。在这里，人位于一个由id和名称组成的对象上。所以这两个函数都是正确的

interface Person {
  id?:number;
  name: string;
}
const t = (person: Person) => `id: ${person.id} name:${person.name}`

// Equivalent to 

const t2 = ({id, name}: Person) => `id: ${id} name:${name}`

as

用于铸造

class Employee {
  name:string;
  salary: number;
}
const emp = new Employee()

console.log(emp.id) //Error here since emp does not have id.
console.log((emp as Person).id) //NO error

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编辑此问题的答案并将其修复：

const t=（{id，name}:{id:number，name:string}）=>`id:${id}name:${name}`

希望下面的内容能帮助你，给你一个想法

 const t = ({id, name}) =>  `id: ${id} name:${name}`;


 interface tParams {
     id: number,
     name: string
 }

 const t2 = ({id, name} : tParams) => `id: ${id} name:${name}`;

 //testing code below
 let tParamsObject: tParams = {
     id: 123,
     name: "James"
 };

 console.log (t(tParamsObject));
 console.log (t2(tParamsObject));

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这是干什么用的？还是用错别字。。这是针对泛型的。。这里不用