Typescript 如何从接口键和键值创建字符串文字类型

有可能吗？对你应该这样做吗？可能不是...

以下解决方案将生成对给定Payload类型有效的字符串文字类型的所有可能组合.

type Expand<T> = T extends infer U ? { [K in keyof U]: U[K] } : never

type OptionalProps<T extends object> = Exclude<{
  [K in keyof T]: T extends Record<K, T[K]>
    ? never
    : K
}[keyof T], undefined>

type AllPermutations<T extends string, O extends string> = [T] extends [keyof T] ? [] : {
  [K in T]: 
    K extends O 
      ? [
          ...(AllPermutations<Exclude<T, K>, O> extends infer U extends string[] 
            ? U 
            : never)
        ]
        | [
            K, ...(AllPermutations<Exclude<T, K>, O> extends infer U extends string[] 
              ? U 
              : never)
          ]
      : [
          K, ...(AllPermutations<Exclude<T, K>, O> extends infer U extends string[] 
            ? U 
            : never)
        ]
}[T]

type Join<T extends string[]> = 
  T extends [infer L extends string, ...infer R extends string[]]
    ? `${L}${Join<R>}`
    : ""

type CreatePayloadString<T extends string[], P extends Record<string, any>> = 
  T extends [infer L extends string, ...infer R extends string[]]
    ? `<${L}>${P[L & keyof P]}</${L}>${CreatePayloadString<R, P>}` 
    : ""

type CondensedPayload = {
  [K in AllPermutations<keyof Payload, OptionalProps<Payload>> as Join<K>]: 
    `<payload>${CreatePayloadString<K, Payload>}</payload>`
} extends infer O ? O[keyof O] : never

一些测试:

// Error 
const payload1: CondensedPayload = '<payload><apiVersion>2.0</destination></payload>'
const payload2: CondensedPayload = '<payload><email>asd@asd.com</email></payload>'
const payload3: CondensedPayload = '<payload><apiVersion>2.0</apiVersion><email>asd@asd.com</email></payload>'
const payload4: CondensedPayload = '<payload><apiVersion>abc</apiVersion><destination>some string</destination></payload>'

// Valid
const payload4: CondensedPayload = '<payload><apiVersion>abc</apiVersion><destination>home</destination></payload>'

但说真的，对这样甚至更好的东西使用XML解析器:不要使用XML.

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