Question

我需要使用objects语法合并两个es6数组，并且没有重复项。在我的任务中，对象中没有几个其他按键。

但是我认为这不是问题。在array2中，array1内始终包含对象

这是数组

array1 = [
   {name: 'one', count: 0}, 
   {name: 'two', count: 0}, 
   {name: 'three', count: 0},
   {name: 'four', count: 0} 
]

array2 = [
   {name: 'two', count: 4},
   {name: 'four', count: 2}
]

我需要这个：

array3 = [
  {name: 'one', count: 0},
  {name: 'two', count: 4},
  {name: 'three', count: 0},
  {name: 'four', count: 2}
]

Answer 1

const array1 = [
  {name: 'one', count: 0}, 
  {name: 'two', count: 0}, 
  {name: 'three', count: 0},
  {name: 'four', count: 0} 
]
const array2 = [
  {name: 'two', count: 4},
  {name: 'four', count: 2}
]

const result = array1.map(a => {
  const b = array2.find(b => b.name === a.name);
  if (b) {
    return {name: a.name, count: a.count + b.count}
  }
  return a;
});

console.log(result);

Answer 2

let all = array1.concat(array2)
//avoid dupplicates by  key
let map = all.reduce((acc,item)=>{acc[item.name] = item; return acc;},{});
//convert map back to array
let array3 = Object.values(map)
console.log(array3);
//
[ { name: 'one', count: 0 },
  { name: 'two', count: 4 },
  { name: 'three', count: 0 },
  { name: 'four', count: 2 } ]

Answer 3

map可用于在数组的每个元素上调用函数。与使用find的版本的主要区别在于它将遍历array2的所有元素，因此，如果数组中有2个元素带有name: 'two'的元素，则这两个元素都会使计数。使用find，只有找到的拳头会添加。

const array1 = [
  {name: 'one', count: 0}, 
  {name: 'two', count: 0}, 
  {name: 'three', count: 0},
  {name: 'four', count: 0} 
  ],
  array2 = [
  {name: 'two', count: 4},
  {name: 'four', count: 2}
  ];

const array3 = array1.map(obj => {
  let _count = obj.count;
  array2.forEach(obj2 => {
    if(obj2.name === obj.name){
      _count += obj2.count;
    }
  });
  return { name: obj.name, count: _count };
});
    
console.log(array3);

编辑：根据@HMR建议修改（感谢指出我的错误）+在array2上使用了forEach，因为我不使用任何返回的数组

我如何合并两个数组对象

3 个答案: