Typescript 不会从平面属性函数中推断出解析的属性

Question

我有一个函数负责扁平化对象属性。

async function query<T = {}>(query: string): Promise<T> {
  return Promise.resolve({
    value: 'random value here',
    randomAttribute: 'randomValue'
  }) as any
}

type QueryResult<T> = {
  id: string
  data: T
}

async function runSomeQuery() {
  const result = await query<QueryResult<{ name: string, age: string }>>('SELECT SOMETHING FROM ANOTHER THING')
  const parsedResult = flatResult(result)
  return parsedResult.name
}

function flatResult<T extends QueryResult<{}>>(queryResult: T) {
  return {
    id: queryResult.id,
    ...queryResult.data
  }
}

但是flatResult() 推断的返回类型缺少名称和年龄属性，所以当我尝试执行parsedResult.name 时出现错误。

游乐场链接：https://www.typescriptlang.org/play/?ssl=24&ssc=2&pln=1&pc=1#code/IYZwngdgxgBAZgV2gFwJYHsIwI4IKYBOYAPACowC8MA3gL4B8AFLoWAFwwjIGoQDmASg4AFAugC2qEHjL0aAKBgwCeZAgJZREqXgB0KkOgA2ANzyNqipTBPAj+DgHICwCABMJNu-hgALQniOADRWSi7uEgCCyNyoAEYIyHhO4R7iAGregVa0AjCg+RBg8rTy8shgAA54MACK+EQASnggCEbIspQKSqhuHFw8-FZuwMjAHKQlZaCQsIgoGFgESADKEnj1rIx5lkpQmFzKLW3IXcAA7sCopyxExJtNx+3E1DAQwOLJnLH8Qfl8XwGvD4MAYTEcKwAogAZSEAYXIKwA8gBZSGkAASAEkAHIAcRgADFGqiYJEcUjMZDGjBMbi8Y4BFZ9hBDpVgARpG5mq12l04EZRjyTowDCcmWFVOosOzOXhuU9kLp3p8pvJ5lA0Jh4ILkMLnuQ8AAPJLuEB1BpgfUdOj0Ji3K2KiY7KwqNQabrWXocB3W3S9ELWGC6EO+xW6EZjHIlIA

那么，我应该怎么做才能让这个推断正确？

Answer 1

您可以简单地依赖 TypeScript 泛型中的类型参数干扰：即让传递给 flatResult() 的参数确定返回类型：

function flatResult<T>(queryResult: QueryResult<T>) {
  return {
    id: queryResult.id,
    ...queryResult.data
  }
}

之所以如此，是因为编译器会查看参数queryResult 的类型。当您在此行中调用/调用flatResult 时：

// In the line below, `result` has the correct inferred type
const result = await query<QueryResult<{ name: string, age: string }>>('SELECT SOMETHING FROM ANOTHER THING')

// And this type will be passed into `flatResult()` and available to the generic type
const parsedResult = flatResult(result)

您实际上已经通知flatResult result 的类型应该是任何类型，键入为QueryResult<{ name: string, age: string }>。

See proof-of-concept example on TypeScript Playground.