TypeScript：从嵌套类型获取类型

huangapple
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英文:

TypeScript: Get type from a nested type

问题

I have a type (Note: Both a and foo can be optional):

  1. type Something = {
  2.   a?: {
  3.     foo?: {
  4.       bar: {
  5.         c: {
  6.           id: string,
  7.           countryCode: number,
  8.           animal: { ... }
  9.         }
  10.       }
  11.     }
  12.   }
  13. }

Now I want to have another type MyType, which is bar in the example above. So MyType will be:

  1. type MyType = {
  2.   c: {
  3.     id: string,
  4.     countryCode: number,
  5.     animal: { ... }
  6.   }
  7. }

My question: How could I create MyType from type Something?

Update: See this comment for the answer: Stack Overflow Comment

英文:

I have a type (Note: Both a and foo can be optional):

  1. type Something = {
  2.   a?: {
  3.     foo?: {
  4.       bar: {
  5.         c: {
  6.           id: string,
  7.           countryCode: number,
  8.           animal: { ... }
  9.         }
  10.       }
  11.     }
  12.   }
  13. }

Now I want to have another type MyType, which is bar in the example above. So MyType will be:

  1. type MyType = {
  2.   c: {
  3.     id: string,
  4.     countryCode: number,
  5.     animal: { ... }
  6.   }
  7. }

My question: How could I create MyType from type Something?

Update: See this comment for the answer: https://stackoverflow.com/questions/76333795/typescript-get-type-from-a-nested-type?noredirect=1#comment134607414_76333795

答案1

得分: 1

The easiest way would be:
type MyType = NonNullable<NonNullable<Something["a"]>["foo"]>["bar"];

However, I can suggest a more dynamic approach as well:

  1. type Something = {
  2.   a?: {
  3.     foo?: {
  4.       bar: {
  5.         c: {
  6.           id: string;
  7.           countryCode: number;
  8.         };
  9.       };
  10.     };
  11.   };
  12. };
  14. type RetrieveSubProperty<T, K extends string[]> = [] extends K
  15.   ? T
  16.   : K extends [infer First extends keyof T, ...infer Rest extends string[]]
  17.   ? RetrieveSubProperty<NonNullable<T[First]>, Rest>
  18.   : never;
  20. type Bar = RetrieveSubProperty<Something, ["a", "foo", "bar"]>;

playground

英文:

The easiest way would be:
type MyType = NonNullable&lt;NonNullable&lt;Something[&quot;a&quot;]&gt;[&quot;foo&quot;]&gt;[&quot;bar&quot;];

However, I can suggest a more dynamic approach as well:

  1. type Something = {
  2.   a?: {
  3.     foo?: {
  4.       bar: {
  5.         c: {
  6.           id: string;
  7.           countryCode: number;
  8.         };
  9.       };
  10.     };
  11.   };
  12. };
  14. type RetrieveSubProperty&lt;T, K extends string[]&gt; = [] extends K
  15.   ? T
  16.   : K extends [infer First extends keyof T, ...infer Rest extends string[]]
  17.   ? RetrieveSubProperty&lt;NonNullable&lt;T[First]&gt;, Rest&gt;
  18.   : never;
  20. type Bar = RetrieveSubProperty&lt;Something, [&quot;a&quot;, &quot;foo&quot;, &quot;bar&quot;]&gt;;

playground

huangapple
  • 本文由 发表于 2023年5月25日 23:07:15
  • 转载请务必保留本文链接：https://go.coder-hub.com/76333795.html
  • nested
  • types
  • typescript
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