ES6 Object Destructuring Default Parameters

I'm trying to figure out if there's a way to use object destructuring of default parameters without worrying about the object being partially defined. Consider the following:

(function test({a, b} = {a: "foo", b: "bar"}) {
  console.log(a + " " + b);
})();

When I call this with {a: "qux"}, for instance, I see qux undefined in the console when what I really want is qux bar. Is there a way to achieve this without manually checking all the object's properties?

Answers:

Answer

Yes. You can use "defaults" in destructuring as well:

(function test({a = "foo", b = "bar"} = {}) {
  console.log(a + " " + b);
})();

This is not restricted to function parameters, but works in every destructuring expression.

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.