CODEWITHSUNDEEP
javascriptarraysecmascript-6destructuring
tres.14159
tres.14159

Reputation: 930

Array destructuring with a ternary operator

I am trying to concat (with uniques values) two arrays and if the second array sometimes is a string.

Maybe it have a bug, but these are my three tryings:

let a = 'abcdefg'
// First try
[...new Set([...[], ...(typeof(a) == 'string'? [a]: a))]
// Second try
[...new Set([...[], [(typeof(a) == 'string'? ...[a]: ...a)]]
// Third try
[...new Set([...[], (typeof(a) == 'string'? ...[a]: ...a)]

Upvotes: 2

Views: 1995

Answers (3)

Me.Name
Me.Name

Reputation: 12544

If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]

let a = 'abcdefg'

const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];

console.log(createArr(a));
console.log(createArr([a,a,'aa']));

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386728

Instead of 

[...new Set([...[], ...(typeof a === 'string' ? [a] : a))]

take, watch the round, square, round and squere closing brackets at the end.

[...new Set([...[], ...(typeof a === 'string' ? [a] : a)])]
//                                                      ^


let a = 'abcdefg'

console.log([...new Set([...[], ...(typeof a === 'string' ? [a] : a)])]);

Upvotes: 4

Ori Drori
Ori Drori

Reputation: 192422

Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:

const a = 'abcdefg'
console.log([...new Set([].concat([], a))])
console.log([...new Set([].concat([], [a]))])

Upvotes: 2

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