TS - 在不获取联合类型的情况下推断键

Question

here is a playground:

https://www.typescriptlang.org/play?target=9#code/C4TwDgpgBAjlC8UDeBYAUFKBtA1hEAXFAM7ABOAlgHYDmAukahplAPYBGAVgIxECGVEABp0LNlwBM-QWPQBfOenQBjVlVKwArgmSjYU3c0wceRAOTAIpMyKPjOBi1eASzstHNuYYAZkZ7jLl4oCS8WEwMJHwAWdwUlNBU1DRhNCDIQADEqHQAeABUoCAAPSyoAE2JYAD4AChMifIBKBGqoXIBpItKICqq8EFYAMyh8upgiDqbGrA66LDMTbjM6VsNMVXVgZCWhewk5HRMsGDo9MghgTTIcpfkEjeTt1LIdVPSsqlrUpoTNjWIrAAthBWMAABbpIY5RC1PhkGhEAAK8L4IMsZGIuVAkGGWjI1SwAAY6C14G0mJgADaXKAXYhEABKl2uVHy4Ag2I5eJeXNxI3hNGq1XOVje1zhCN+zAuVxudKs90SaH+23pOkBILBkLI0NqZl8ZhaAHpjVAaBQAG7UGhQEFQPhQTRUChqKDUUgQPjlNgjahDdKUWhQVRkC7KbY4iAJdBAA

Basically having a function where I can send in an argument that returns some value. It gives me a union however since there are keys with the same name that have different types:

const qu = {
  q2: {
    obj1: 'test',
    obj2: 'test2'    
},
  q3: {
    obj1: 2,
    obj2: 234    
}}

is there a way to infer the type? I've tried with conditionals but with no luck. Or is this a TS limitation?

Here is full code example:

type q = {
  [key: string]: {
    obj1: any,
    obj2: any    
}}

const qu = {
  q2: {
    obj1: 'test',
    obj2: 'test2'    
},
  q3: {
    obj1: 2,
    obj2: 234    
}}


const queryFn = <T extends q>(obj: T) => <K extends keyof T>(q: K): T[K]['obj1'] => {
  const {obj1, obj2} = obj[q]
  return obj1
}

  const qur = queryFn(qu)

const someotherfn = (arg: Parameters<typeof qur>[0]) => {
  let res: ReturnType<typeof qur<typeof arg>>
  res = qur(arg)
  return res
}

const res = someotherfn('q3') // giving me a union instead of inferring correct type

Answer 1

The problem here is that someotherfn is not itself generic ; if you don't explicitly annotate the call signature as generic, then the return type someotherfn will not depend on its input type.

The fix could look like this:

const someotherfn = <K extends Parameters<typeof qur>[0]>(arg: K) => {
  let res: ReturnType<typeof qur<K>>
  res = qur(arg)
  return res
}

where we just make the function generic in K , the type of arg . Now things behave as expected:

const res = someotherfn('q3');
// const res: number

Playground link to code