通过 if 条件从 `Array.flatMap()` 中不返回任何元素

Question

I have this code:

  const myFunc = function (t) {
    return myArray.flatMap(clip =>
      (t < clip.start || t < clip.end) ? // Valid objects are returned in this *if* condition
        [
          { time: clip.start },
          { time: clip.end }
        ] : // how to return nothing in this *else* condition. Absolutely nothing?
        [
          {  },
          {  }
        ]
    )
  }

The above code used a ternary operator of condition? exprIfTrue: exprIfFalse condition? exprIfTrue: exprIfFalse .

Currently I'm returning empty objects of { } in the case of exprIfFalse .

How can I return nothing in the case of exprIfFalse ? I mean, I want absolutely nothing. I mean no array element.

Answer 1

Why cant you just return an empty array, any how Array.flat will remove those empty array from final code. In your case the array is not empty as [] , its an array with two empyty objects as [{}, {}] that will produce two empty objects {}, {} in the final output after Array.flat

You have to return something from flatMap . If you return nothing, the corresponding nodes will be added as undefined . That wont be removed with Array.flat . Best option is to return an empty array as below.

Pseudo Code

 const myArray = [1, 2, 3, 4, 5]; const myFunc = function (t) { return myArray.flatMap(clip => (clip % 2 === 0)? // Valid objects are returned in this *if* condition [ { value: clip }, { value: clip } ]: // how to return nothing in this *else* condition. Absolutely nothing? [] ) } console.log(myFunc());