Use optional property of an interface as a type ( in openapi-typescript)

Question

How can I use a property of an interface as a type for a variable in typescript ??
Here I want to access the property: string type and use it as a type for a variable but I cannot access it.

interface foo {
  bar?: {
    baz: {
      property: string;
    };
  };
}

function f(input: foo['bar']['baz']['property']) {
  console.log(input);
}

I was trying to find some optional chaining rule for this, but none of the JavaScript chaining methods worked here.

Error

Property 'baz' does not exist on type '{ baz: { property: string; } ' | undefined

artem · Accepted Answer · 2023-01-08 12:45:57Z

5

To remove | undefined part from the foo['bar'] type you can use built-in NonNullable type, then you can access its properties:

function f(input: NonNullable<foo['bar']>['baz']['property']) {
  console.log(input);
}

answered Jan 8, 2023 at 12:45

artem

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