Javascript 删除有条件的重复对象

发布于05月17日

假设我有一个这样的对象列表:

let b = [
  {
    name: "test1", 
    connectedTo: "",
  },
  {
    name: "test1",
    connectedTo: "test1.test2.test3"
  },
  {
    name: "test2",
    connectedTo: "",
  },
  {
    name: "test3",
    connectedTo: "",
  }
]

我想得到name个没有重复的元素，如果有重复的元素，也要得到一个没有空connectedTo的元素.从上面的例子来看，我期望的结果是:

let result = [
  {
    name: "test1",
    connectedTo: "test1.test2.test3"
  },
  {
    name: "test2",
    connectedTo: "",
  },
  {
    name: "test3",
    connectedTo: "",
  }
]

推荐答案

下面是一种利用Array#reduce函数的方法:

let b = [
  {
    name: "test1", 
    connectedTo: "",
  },
  {
    name: "test1",
    connectedTo: "test1.test2.test3"
  },
  {
    name: "test2",
    connectedTo: "",
  },
  {
    name: "test3",
    connectedTo: "test1.test2",
  },
  {
    name: "test3",
    connectedTo: "",
  }
];

const result = Object.values(b.reduce((acc, cur) => {
  if(!acc[cur.name] || !acc[cur.name].connectedTo)
    acc[cur.name] = cur;
  return acc;
}, {}));

console.log(result);