Add key in Array of object if match condition

Question

I am trying to get an array of objects based on a condition. If the value is true then only add that key and value to new array of object.

But, It gives error:

const arr = [{id: 123, val: 'abcd', other: 'abcd'}, {id: 123, val: 'abcd', other: 'abcd', show: true}, {id: 123, val: 'abcd', other: 'abcd', show: false}, {id: 123, val: 'abcd', other: 'abcd'}]

const newArr = arr.map(item => ({id: item.id, val: item.val != null ? item.val : ''}))

console.log(newArr); //It is working

const arr = [{id: 123, val: 'abcd', other: 'abcd'}, {id: 123, val: 'abcd', other: 'abcd', show: true}, {id: 123, val: 'abcd', other: 'abcd', show: false}, {id: 123, val: 'abcd', other: 'abcd'}]

const expectedArr = arr.map(item => ({id: item.id, val: item.val != null ? item.val : '', (item.show) && (show: item.show)}))

console.log(expectedArr); //Shows error

Expected Result:

[
  {
    "id": 123,
    "val": "abcd"
  },
  {
    "id": 123,
    "val": "abcd"
    "show": true
  },
  {
    "id": 123,
    "val": "abcd"
  },
  {
    "id": 123,
    "val": "abcd"
  }
]

Any help would be greatly appreciated.

(show) && (show: val.show)}) should be (item.show) && (show: item.show)})? — lissettdm
– lissettdm, Commented Oct 16, 2020 at 12:42
Change (item.show) && (show: item.show) to ...(item.show && {show: item.show}) — Yousaf
– Yousaf, Commented Oct 16, 2020 at 12:48

Mister Jojo · Accepted Answer · 2020-10-16 15:51:27Z

3

that ?

const arr = 
      [ { id: 123, val: 'abcd', other: 'abcd' } 
      , { id: 123, val: 'abcd', other: 'abcd', show: true  } 
      , { id: 123, val: 'abcd', other: 'abcd', show: false } 
      , { id: 123, val: 'abcd', other: 'abcd' } 
      ] 

const newArr = arr.map(({id,val,show})=>show?{id,val,show}:{id,val})

console.log( newArr )

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Oct 16, 2020 at 15:51

answered Oct 16, 2020 at 12:53

Mister Jojo

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Comments

Patrick Bateman · Accepted Answer · 2020-10-16 12:59:01Z

2

You can rewrite it like this. with this syntax you, when item.show is true the parenthesis resolve to and object and then you just spread it into your object. You can also use nullish coalescing to assign item.val

 const expectedArr = arr.map((item) => ({
      id: item.id,
      val: item.val ?? "",
      ...(item.show && { show: item.show }),
    }));

answered Oct 16, 2020 at 12:59

Patrick Bateman

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Comments

Liam · Accepted Answer · 2020-10-16 12:48:29Z

1

Currently, it contains multiple conditions on one line inside Array.map function and it's not valid.

const arr = [{id: 123, val: 'abcd', other: 'abcd'}, {id: 123, val: 'abcd', other: 'abcd', show: true}, {id: 123, val: 'abcd', other: 'abcd', show: false}, {id: 123, val: 'abcd', other: 'abcd'}]

const result = arr.map(({ id, val, show }) => {
  const newObj = {
    id,
    val: val != null ? val : ''
  };
  if (show) {
    newObj['show'] = show;
  }
  return newObj;
});

console.log(result);

edited Oct 16, 2020 at 12:48

Liam

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answered Oct 16, 2020 at 12:45

Derek Wang

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Comments

The Bomb Squad · Accepted Answer · 2020-10-16 14:40:54Z

const arr = [{id: 123, val: 'abcd', other: 'abcd'}, {id: 123, val: 'abcd', other: 'abcd', show: true}, {id: 123, val: 'abcd', other: 'abcd', show: false}, {id: 123, val: 'abcd', other: 'abcd'}]

const expectedArr = arr.map(item => ({id: item.id, val: item.val? item.val: '', show: item.show? item.show: 0}))

console.log(expectedArr); //solved-ish

//but honestly, that's the furthest i can go with your strange code golfed syntax setup.. to be honest, using ? is similar to if but NOT if, it's closer to || but anyway

console.log('skip what was above, the moment of truth below')

const theRealAnswer = arr.map(item=>{ const i={id: item.id, val: item.val? item.val: '', show: item.show? item.show: 0};if(!i.show){delete(i.show)}return(i) })

console.log(theRealAnswer)

//now, the logical construct(what u gave the mapper) was impossible to do what u wanted. coding isn't just syntax, it's logic too. you cannot assign a key to completely delete the key itself

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