javascript - typescript类型推断，如何根据object的一个key推断对象其他属性？

Question

先上代码：

type DetailDataType = {    detailType: 'type1'    params: {        a1: string        test1: number    }} | {    detailType: 'type2'    params: {        ss: Array<string>        aa: Record<string, boolean>    }}type dataType = { id: string } & DetailDataTypeconst testFn = <Type extends dataType['detailType']>(type: Type, data: unknown) => {    if (type === 'type1') {        const { a1, test1 } = data;    }    if (type === 'type2') {        const { ss, aa } = data;    }};testFn('type1', { a1: '', test1: 1 });testFn('type1', { ss: [], aa: { test: false } });

问题：
该如何定义函数testFn的data参数，才能让data的解构与DetailDataType的定义相符，并且编辑器有提示，现在data的解构全是any，如下：

求ts大神解答，谢谢！！！

Answer 1

别拆开，判断detailType的值来收窄类型

const testFn = (data: dataType) => {  if (data.detailType === 'type1') {      const { a1, test1 } = data.params;  } else if (data.detailType === 'type2') {      const { ss, aa } = data.params;  }};

Answer 2

type DetailDataType =  | {      detailType: "type1";      params: {        a1: string;        test1: number;      };    }  | {      detailType: "type2";      params: {        ss: string[];        aa: Record<string, boolean>;      };    };type dataType = { id: string } & DetailDataType;type Params<Type extends dataType["detailType"]> = Extract<  dataType,  { detailType: Type }>["params"];const testFn = <Type extends dataType["detailType"]>(  type: Type,  params: Params<Type>,) => {  if (type === "type1") {    const { a1, test1 } = params as Params<"type1">;  }  if (type === "type2") {    const { ss, aa } = params as Params<"type2">;  }};testFn("type1", { a1: "", test1: 1 });testFn("type2", { ss: [], aa: { test: false } });