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How to transform union of similar objects to object type using TypeScript typings?

Input

type I = {
  key: 'foo',
  value: Foo,
} | {
  key: 'bar',
  value: Bar,
};

Output

type O = {
  foo: Foo,
  bar: Bar,
};

I am not sure it is possible. But who knows? Note that the task is not a duplicate to that.

How to transform union of similar objects to object type using TypeScript typings?

Input

type I = {
  key: 'foo',
  value: Foo,
} | {
  key: 'bar',
  value: Bar,
};

Output

type O = {
  foo: Foo,
  bar: Bar,
};

I am not sure it is possible. But who knows? Note that the task is not a duplicate to that.

• javascript
• typescript

Share Improve this question Follow edited Dec 20, 2020 at 1:18 jcalz 333k2929 gold badges443443 silver badges442442 bronze badges asked Dec 19, 2020 at 23:52 avdotionavdotion 13711 silver badge1010 bronze badges 2

• 3 That's not an intersection. – Jared Smith Commented Dec 20, 2020 at 0:01
• It's not clear what the algorithm would even be to get from your "input" to your "output" unless you had a type that happened to be a union of a bunch of types with key properties of type string and value properties of various types. And how would you use this output? You want an expression that produces O from I? How does that help you? O is not that hard to type into your code -- you did it above. So yeah, see whether you can improve or clarify the question, I'd say. – David P. Caldwell Commented Dec 20, 2020 at 0:07

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 6

In TypeScript 4.1 this is a straightforward mapped type with key remapping. You can iterate over each member T of the I union (it's a union, not an intersection) and look up the key/value types: T['key'] for the key and T['value'] for the value. Like this:

type O = { [T in I as T['key']]: T['value'] };
/* type O = {
    foo: Foo;
    bar: Bar;
} */

Playground link to code

How to transform union of similar objects to object type using TypeScript typings?

Input

type I = {
  key: 'foo',
  value: Foo,
} | {
  key: 'bar',
  value: Bar,
};

Output

type O = {
  foo: Foo,
  bar: Bar,
};

I am not sure it is possible. But who knows? Note that the task is not a duplicate to that.

How to transform union of similar objects to object type using TypeScript typings?

Input

type I = {
  key: 'foo',
  value: Foo,
} | {
  key: 'bar',
  value: Bar,
};

Output

type O = {
  foo: Foo,
  bar: Bar,
};

I am not sure it is possible. But who knows? Note that the task is not a duplicate to that.

• javascript
• typescript

Share Improve this question Follow edited Dec 20, 2020 at 1:18 jcalz 333k2929 gold badges443443 silver badges442442 bronze badges asked Dec 19, 2020 at 23:52 avdotionavdotion 13711 silver badge1010 bronze badges 2

• 3 That's not an intersection. – Jared Smith Commented Dec 20, 2020 at 0:01
• It's not clear what the algorithm would even be to get from your "input" to your "output" unless you had a type that happened to be a union of a bunch of types with key properties of type string and value properties of various types. And how would you use this output? You want an expression that produces O from I? How does that help you? O is not that hard to type into your code -- you did it above. So yeah, see whether you can improve or clarify the question, I'd say. – David P. Caldwell Commented Dec 20, 2020 at 0:07

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 6

In TypeScript 4.1 this is a straightforward mapped type with key remapping. You can iterate over each member T of the I union (it's a union, not an intersection) and look up the key/value types: T['key'] for the key and T['value'] for the value. Like this:

type O = { [T in I as T['key']]: T['value'] };
/* type O = {
    foo: Foo;
    bar: Bar;
} */

Playground link to code

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