在打字稿中将嵌套对象叶键值声明为未定义 [英] Declare nested object leaf key values as undefined in typescript

问题描述

考虑如下嵌套对象

class Person {
    name: string = 'name';
    address: {street: string, pincode: string} = {street: 'street', pincode: '44555'};
}

有一个实用函数(undefinedAllLeadNodes) 可以取消定义所有叶节点.所以传递一个新的 Person() 在下面创建

There is a utility function(undefinedAllLeadNodes) which undefines all leaf nodes. So passing a new Person() creates below

person.name = undefined
person.address.street = undefined;
person.address.pincode = undefined;

如果我声明如下

export function undefineAllLeafProperties<
    T extends object
>(obj : T) :  {[key: string] : any} {
    ...undefine all leaf nodes here
}

它显示了 person.address.street 的编译错误，对于 {[key: string] : any} 返回类型声明听起来不错.

It shows compilation error for person.address.street and it sounds fine for {[key: string] : any} return type declaration.

但是什么是效用函数的正确返回类型(undefinedAllLeadNodes)这样我就可以像下面一样访问它而不会出现任何错误

But What can be the proper return type for the utility function(undefinedAllLeadNodes) so that i can access it like below without any errors

person.address.street = 'update only street';

推荐答案

你可以定义这样的类型:

You may define such a type like this:

type Undefine<T extends object> = {
  [K in keyof T]: T[K] extends object ? Undefine<T[K]> : T[K] | undefined
}

游乐场链接

这里我们使用映射类型进行迭代所有键(使用 keyof 运算符)提供的 T 对象和 条件类型检查T[K]的值是否(使用查找类型) 又是一个对象.当它是我们递归迭代 T[K] 键.否则(当 T[K] 不是对象时)我们只需更改它的类型以允许沿 T[K]undefined 值>.

Here we're using mapped types to iterate over all keys (using keyof operator) of the provided T object and conditional type to check whether the value of T[K] (using lookup type) is an object again. And when it is we recursivly iterate over the T[K] keys. Otherwise (when T[K] is not an object) we just change it's type to allow undefined value along the main type of T[K].

请记住，在 undefineAllLeafProperties 函数代码中，您可能必须键入 assert return values .因为打字稿通常无法推断条件类型，当它依赖于未指定的泛型类型参数时.

Keep in mind that inside the undefineAllLeafProperties function code you may have to type assert return values . Because typescript generally cannot infer the exact shape of the conditional type when it depends on unspecified generic type parameter.

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