Question

假设我有两个对象object_one

{
  1: {item1 /*updated*/}, //note updated represents more recent item1 data
  2: {item1 /*updated*/},
  3: {item1 /*updated*/}
}

和object_two

{
  1: {item1, item2, item3},
  2: {item1, item2, item3},
  4: {item1, item2, item3},
  5: {item1, item2, item3}
}

我想将object_one添加到object_two，添加object_two没有的任何元素。我还想使用{1}}版本的item1并将其更新为object_two＆＃39; s

的值

期望的结果

object_one

我自己尝试过这样做，但我的解决方案是手动的，并且不适用于{ 1: {item1 /*updated*/, item2, item3}, 2: {item1 /*updated*/, item2, item3}, 3: {item1 /*updated*/}, 4: {item1, item2, item3}, 5: {item1, item2, item3} }和object_one的所有长度。我们非常感谢某些方向

Answer 1

有关详细信息，请参阅Object.entries和Object.assign。

// Original.
const original = {
  1: {A: '1A', B: '1B', C: '1C'},
  2: {A: '2A', B: '2B', C: '2C'},
  4: {A: '4A', B: '4B', C: '4C'},
  5: {A: '5A', B: '5B', C: '5C'}
}

// Merge.
const update = {
  1: {A: '1A updated'},
  2: {B: '2B updated'},
  3: {C: '3C updated'}
}

// Combine.
const merge = (original, update) => {
  const x = {...original}
  Object.entries(update).forEach(([key, value]) => {
    x[key] = Object.assign({}, x[key], value)
  })
  return x
}

// Output.
const output = merge(original, update)

// Proof.
console.log(output)

结合数组和插入差异

1 个答案: