Spread syntax ES6 with statement

Question

I tried to write ternary operator with spread syntax and copy two objects. Is it possible to use ternary operator with spread syntax inside with literal objects? My code works okay, I just want to optimize it.

hintStyle: disabled ? {...globalStyles.hint, ...globalStyles.hintDisabled} : globalStyles.hint,

I want to write like this:

hintStyle: {...globalStyles.hint, {disabled ? ...globalStyles.hintDisabled : {}}},

Aerial

Does it work? If not, do you get errors? If so, what errors? What's the question you're asking? – Bugaboo

My first code works , how to use spread es6 operator with ternary operator inside. The error is 'Unexpected token' – Aerial

don't you need a spread operator before {disabled ? ...globalStyles.hintDisabled : {}} – Schreck

Rest spread is transpiled to Object.assign invocation, which certainly can be written in form with conditional argument applying – Young

Answer 1

Spread is not an operator, it's part of the object literal syntax (or at least it will be when the proposal is accepted). You need to write

{...globalStyles.hint, ...(disabled ? globalStyles.hintDisabled : {})},

Hominy

What if you only want to spread in one case bot not in the other? Lets say string vs array is the check. – Smutch

@TheFool You can only spread an object into an object. If you have a value that is not an object, how do you want to make it part of the object? – Hominy

you can spread one array into another or you can just add another item – Smutch

@TheFool This question is not about arrays though. Do you use array spread syntax? – Hominy

Answer 2

I ran into this same problem today, my use case was simple enough that I could do this:

// given any-typed parameters a and b, I want to append b to a
// if a is iterable, I want to use spread.
// Initially I had:

const fn1 = (a, b) => [Symbol.iterator in a ? ...a : a, b]

// This was my solution:

const fn2 = (a, b) => Symbol.iterator in a ? [...a, b] : [a, b];

Punic