Can the 'spread operator' replace an object with same uuid? Solution: Object.assing()

Question

I was trying to replace a property of an Object in an array with the spread syntax like this:

const origArray = [
  {
    "uuid":"first-team-uuid",
    "name":"Team 1",
    "players": [
      "first-team-uuid"
    ]
  },
  {
    "uuid":"second-team-uuid",
    "name":"Team 2",
    "players":[]
  }
]

const doesNotWork = (prev, index, newName) => [...prev, {...prev[index], name: newName}]

const result1 = doesNotWork(origArray, 0, "Team3")

console.log(result1)

// # I know this works:

const doesWork = (prev, index, newName) => {
  let old = [...prev]
  old.splice(index, 1, {...prev[index], name: newName});
  return old;
}

const result2 = doesWork(origArray, 0, "Team3")

console.log(result2)

I expect reslut1 to be like result2, but I seem to be wrong. I would like to write this in a singe line function and not with the workaround I currently have, if possible.

Answer 1

I can suggest these ways, through a filter, a map and an object.

But filter way changes the order of elements in array

const origArray = [
  {"uuid":"c752cf08","name":"Team 1",},
  {"uuid":"d46829db","name":"Team 2",},
  {"uuid":"d46829d0","name":"Team 3",}];

const match = 1;
const name = 'Team 100';

//------------------------
const workWithFilter = (prev) =>  
  [...prev.filter((_, i) => i !== match), { ...prev[match], name }];
  
const result1 = workWithFilter(origArray);
console.log('workWithFilter:', result1);

//------------------------
const workWithMap = (prev) => 
  prev.map((v, i) =>  (i === match) ? { ...v, name } : v);

const result3 = workWithMap(origArray);
console.log('workWithMap:', result3);

//------------------------
const workWithObject = (prev) => 
  Object.assign([...prev], { [match]: { ...prev[match], name } });

const result4 = workWithObject(origArray);
console.log('workWithObject:', result4);

//------------------------
const doesWork = (prev) => {
  let old = [...prev]
  old.splice(match, 1, { ...prev[match], name });
  return old;
}

const result2 = doesWork(origArray);
console.log('doesWork:', result2);

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