Typescript extend two interfaces like type?

Question

After reviewing the documentation and this SO Question it looks like the behavior I am looking for is the following:

type A = {
  a:string
};
type B = {
  b:string
};

type C = A & B & {
   c:string
};

My question is: Can the same behavior be done with interfaces in a single line?

I vaguely recall seeing a duplicate out there on this question, but I can't find it. — BobtheMagicMoose
– BobtheMagicMoose, Commented Jan 26, 2021 at 16:59
Please consider giving a minimal reproducible example of what you're trying to do, with enough structure to distinguish good answers from bad answers. Right now it's a bit too minimal: you've got empty object types, so Primary1 and Primary2 are equivalent to each other and Secondary is equivalent to just {prop: string}. You can extend multiple interfaces like interface Secondary extends Primary1, Primary2 { prop: string }, but I'm not sure if that actually meets whatever use case you have without a little more in the example. — jcalz
– jcalz, Commented Jan 26, 2021 at 17:02

jcalz · Accepted Answer · 2021-01-26 17:07:11Z

4

You can extend a comma-separated list of interfaces at once, as long as the interfaces don't have overlapping properties of differing types:

interface Primary1 { prim1: string };
interface Primary2 { prim2: string };
interface Secondary extends Primary1, Primary2 { prop: string }

declare const secondary: Secondary;
secondary.prim1.toUpperCase(); // ok
secondary.prim2.toUpperCase(); // ok
secondary.prop.toUpperCase(); // ok

Playground link to code

answered Jan 26, 2021 at 17:07

jcalz

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