如何动态引用 TypeScript 接口属性的值？

huangapple
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英文:

How to dynamically refer to TypeScript interface property's value?

问题

请看这个愚蠢的例子：

  1. interface Building {
  2.   rooms: number
  3.   kitchen: {
  4.     hasOven: true
  5.   }
  6.   bathroom: {
  7.     numberOfSinks: number
  8.   }
  9. }
  11. interface Select<T> {
  12.   name: keyof T
  13.   fields: (keyof T[this.name])[] // <-- 这不是有效的代码。我应该怎么做？
  14. }
  16. const x: Select<Building> = {
  17.   name: 'kitchen',
  18.   fields: ['hasOven'] // 因为我指定了 'kitchen'，所以它只允许 'hasOven'
  19. }
  21. const y: Select<Building> = {
  22.   name: 'bathroom',
  23.   fields: ['numberOfSinks'] // 因为我指定了 'bathroom'，所以它只允许 'numberOfSinks'
  24. }

上面的代码不起作用。

如何使 Select<T> 接口能够动态引用 name 属性的值，并使用它来确定可用的键值？

非常感谢！

英文:

Consider this silly example:

  1. interface Building {
  2.   rooms: number
  3.   kitchen: {
  4.     hasOven: true
  5.   }
  6.   bathroom: {
  7.     numberOfSinks: number
  8.   }
  9. }
  11. interface Select<T> {
  12.   name: keyof T
  13.   fields: (keyof T[this.name])[] // <-- This is not valid code. What should I do instead?
  14. }
  16. const x: Select<Building> = {
  17.   name: 'kitchen',
  18.   fields: ['hasOven'] // Because I specified 'kitchen', it only allows 'hasOven'
  19. }
  21. const y: Select<Building> = {
  22.   name: 'bathroom',
  23.   fields: ['numberOfSinks'] // Because I specified 'bathroom', it only allows 'numberOfSinks'
  24. }

The above code does not work.

How do I make the Select<T> interface to be able to refer to the name property's value dynamically and use that to determine the available key values?

Thanks so much in advance!

答案1

得分: 2

以下是代码的翻译部分：

  1. 无法在不引入第二个泛型类型的情况下完成：
  3. interface Select<T, K extends keyof T> {
  4.     name: K
  5.     fields: (keyof T[K])[]
  6. }
  8. 所以现在
  10. const a: Select<Building, "bathroom"> = {
  11.     name: 'bathroom',
  12.     fields: ['hasOven'] // 不行！
  13. }
  15. const b: Select<Building, "steering-wheel"> = { // 不行！
  16.     name: 'steering-wheel',
  17.     fields: ['hasOven'] // 不行！
  18. }
  20. const c: Select<Building, "bathroom"> = {
  21.     name: 'steering-wheel', // 不行！
  22.     fields: ['numberOfSinks']
  23. }

Playground链接

英文:

Can't be done without introducing a second generic type:

  1. interface Select&lt;T, K extends keyof T&gt; {
  2.     name: K
  3.     fields: (keyof T[K])[]
  4. }

so now

  1. const a: Select&lt;Building, &quot;bathroom&quot;&gt; = {
  2.     name: &#39;bathroom&#39;,
  3.     fields: [&#39;hasOven&#39;] // NO!
  4. }
  6. const b: Select&lt;Building, &quot;steering-wheel&quot;&gt; = { // NO!
  7.     name: &#39;steering-wheel&#39;,
  8.     fields: [&#39;hasOven&#39;] // NO!
  9. }
  11. const c: Select&lt;Building, &quot;bathroom&quot;&gt; = {
  12.     name: &#39;steering-wheel&#39;, // NO!
  13.     fields: [&#39;numberOfSinks&#39;]
  14. }

Playground Link

huangapple
  • 本文由 发表于 2023年3月10日 02:01:17
  • 转载请务必保留本文链接：https://go.coder-hub.com/75688417.html
  • typescript
  • typescript-typings
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