Destructure Optional Params in Typescript
Matt Gregg
Posted on May 6, 2021
Sometimes you have a function with an optional Object argument that you want to destructure in the function. Like so:
interface SomeObject {
option1: boolean;
stuff: boolean;
}
function foo(param?: SomeObject) {
const { stuff } = param;
}
However you'll get an error because param
could be undefined and TS doesn't like you trying to destructure something that's undefined. There are a couple ways around this...
Define a fallback in the initializer and don't use the ? optional identifier:
function foo(param: SomeObject = {}) {
const { stuff } = param;
}
Use nullish coalescence:
function foo(param?: SomeObject) {
const { stuff } = param ?? {};
}
Or just call the propery on the parameter directly with optional chaining:
function foo(param?: SomeObject) {
anotherFunction(param?.stuff);
}
All of these work and will handle param
being undefined.
💖 💪 🙅 🚩
Matt Gregg
Posted on May 6, 2021
Join Our Newsletter. No Spam, Only the good stuff.
Sign up to receive the latest update from our blog.