如何根据Typescript 中带有泛型的对象嵌套键数组获取数组或对象的正确类型

我首先使用对象键生成建议键，如下所示

const data = {
 content: {
  value: "string", 
 },
 anotherContent: {
  anotherValue: 20
 }
}.

const result = function(["content.value","anotherContent.anotherValue"])

首先，我希望result的型号是{value: number, otherValue: number}.

所以，我这样打字.

<Target, Suggestion extends DotedKeysType<S>[]>(
    target?: Target extends Suggestion ? Target : Suggestion 
  ): {
    [key in Target extends Suggestion ? Target[number] : string]: Target extends Suggestion 
      ? OutputByKeys<S, key>
      : never;
  };

除以下各项外，其他各项运行正常.

• 当目标是content时，我可以按如下方式使用它

<View>{result.content}</View>

• 当它是anotherContent.anotherValue的时候，我必须像这样使用它.

<View>{result["anotherContent.anotherValue"]}</View>

我也意识到，即使我们成功了，我们也会有一个关键的问题

// for a data like this. Both last keys are named 'value'
const data = {content: {value: 10},anotherContent: {value: 20}}

// Keeping the result as an object override the value key in the object.
// Since objects can not have same key as property
const result = function(["content.value","anotherContent.value"]);
result = {value : 20 // `Override value of content with anotherContent`}

但是使用数组的结果是，数组中的每个项可以有不同的名称.例如

result = [myFirstValue, mySecondValue] = function(["content.value","anotherContent.anotherValue"])

其中，myFirstValue的类型必须是content.value的类型，而mySecondValue的类型必须是anotherContent.anotherValue的类型.

到目前为止，这就是我所做的一切.

 <Target, Suggestion extends DotedKeysType<S>[]>(
    target?: Target extends Suggestion ? Target : Suggestion 
  ): OutputByKeys<S, Target extends Sug ? Target[number] : keyof S>[];

这为我提供了目标数组中所有键的联合类型.

result = [value1: string | number, value2: string | number]

Is there a way to get a type like this in case of array?个

// type of `content.value` is `string`
// type of `notherContent.anotherValue` is `number`

const result = function(["content.value","anotherContent.anotherValue"]);

result =>  [myFirstValue, mySecondValue]

result type will be [value1: string, value2: number]

And in case of object where last key are different个

// type of `content.value` is `string`
// type of `anotherContent.anotherValue` is `number`
const result = function(["content.value","anotherContent.anotherValue"])

result => {value: "string", anotherValue: 20}

result type => {value: string, anotherValue: number}

如果可能的话，我不想要十字路口.

// i don't want that if possible
result type => [string & number, string & number]

但我想要这个

result type => [string, number]

我无法预测用户的 Select ，否则我只需将数组类型设置为=> [string, number]，我的问题就解决了.我已经读了很多答案，但我仍然没有找到我需要的.

我以不同的方式要求同样的东西.对象，因为它可能在另一个项目中有用，而数组，因为它是我当前需要的.