typescript-声明子对象的结构
如何将对象的结构声明为其他对象的子对象?我不知道编译时有多少孩子typescript-声明子对象的结构,typescript,Typescript,如何将对象的结构声明为其他对象的子对象?我不知道编译时有多少孩子 let obj = { child1: {id: 1, name: "Name id1"}, child2: {id: 2, name: "Name id2"}, child3: {id: 3, name: "Name id3"}, child4: {id: 4, name: "Name id4"}, child5: {id: 5, name: "Name id5"}, child6
let obj = {
child1: {id: 1, name: "Name id1"},
child2: {id: 2, name: "Name id2"},
child3: {id: 3, name: "Name id3"},
child4: {id: 4, name: "Name id4"},
child5: {id: 5, name: "Name id5"},
child6: {id: 6, name: "Name id6"},
}
所以我想声明类型(或接口){id:number,name:string}
,并且obj
可以为这个新类型的属性拥有任何属性名称和值。感谢@jornsharpe
interface IdName {id: number, name: string}
let list: { [key: string]: IdName } = {};
obj:{[key:string]:…}
?