'spread operator' 可以替换具有相同 uuid 的对象吗？解决方案：Object.assing()

Question

我试图用 spread syntax 替换数组中对象的属性，如下所示：

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)

我希望reslut1 和result2 一样，但我好像错了。如果可能的话，我想用单行函数而不是我目前的解决方法来写这个。

Answer 1

我可以通过过滤器、地图和对象来推荐这些方法。

但是过滤方式改变了数组中元素的顺序

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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'spread operator' 可以替换具有相同 uuid 的对象吗？解决方案：Object.assing()

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

javascript

arrays

json

spread