从 javascript 中的 JSON 对象中删除特定的数组元素

Question

我有以下对象数组-

var myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ]

我想检查数组中是否存在相同术语的多个条目并删除具有 id="" 的节点，这意味着我的输出将具有术语且 id 不等于 ''（空字符串）的条目重复的术语条目。

myarray 逻辑后的最终值如下：

[{"id":"ABC12345","AlphaNumber":"LL8888","terms":"1"},{"id":"","AlphaNumber":"KK6666","terms":"2"},{"id":"ABC12346","AlphaNumber":"RR4444","terms":"3"},{"id":"ABC12347","AlphaNumber":"ASQE223","terms":"5"}]

Answer 1

以terms 作为键创建一个map 对象。如果当前terms在map中不存在或者已经存在的对象没有id，则使用当前对象

const myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ]

const map = {};

for (const o of myarray) {
  if (!map[o.terms]?.id)
    map[o.terms] = o
}

console.log( Object.values(map) )

Answer 2

使用Array#reduce 和Map

1. 将您的数据按terms 分组到Map。
2. 查找id 不是empty 字符串的项目，如果这样的项目不存在，则选择第一个项目。

const 
  myarray = [ { "id": "", "AlphaNumber": "ADF12345", "terms": "1" }, { "id": "ABC12345", "AlphaNumber": "LL8888", "terms": "1" }, { "id": "", "AlphaNumber": "KK6666", "terms": "2" }, { "id": "", "AlphaNumber": "QQ1111", "terms": "3" }, { "id": "ABC12346", "AlphaNumber": "RR4444", "terms": "3" }, { "id": "", "AlphaNumber": "SS1111", "terms": "5" }, { "id": "ABC12347", "AlphaNumber": "ASQE223", "terms": "5" } ],
  
  res = Array.from(myarray.reduce(
    (m, o) => (m.has(o.terms) ? m.get(o.terms).push(o) : m.set(o.terms, [o]), m),
    new Map()
  ).values(), (v) => v.find(o => o.id !== "") || v[0]);

console.log(JSON.stringify(res));

Answer 3

您可以执行 2 个步骤：

1. 按terms分组数据
2. 当组中的数据大于 1 项时过滤数据。否则，将数据保留在只有一项的组中。

if(values.length > 1) values = values.filter(r => r.id !== "");

const myarray=[{"id":"","AlphaNumber":"ADF12345","terms":"1"},{"id":"ABC12345","AlphaNumber":"LL8888","terms":"1"},{"id":"","AlphaNumber":"KK6666","terms":"2"},{"id":"","AlphaNumber":"QQ1111","terms":"3"},{"id":"ABC12346","AlphaNumber":"RR4444","terms":"3"},{"id":"","AlphaNumber":"SS1111","terms":"5"},{"id":"ABC12347","AlphaNumber":"ASQE223","terms":"5"}];
const result = [];

// Step 1
const groupingData = myarray.reduce((acc, curr) => {
  acc[curr.terms] ??= [];
  acc[curr.terms].push(curr);
  return acc;
}, {});

// Step 2
for(let [key, values] of Object.entries(groupingData)){
  if(values.length > 1) values = values.filter(r => r.id !== "");
  result.push(values);
}
console.log(result.flat());

Answer 4

我的建议：

const array = [{
    id: "",
    AlphaNumber: "ADF12345",
    terms: "1"
  }, {
    id: "ABC12345",
    AlphaNumber: "LL8888",
    terms: "1"
  }, {
    id: "",
    AlphaNumber: "KK6666",
    terms: "2"
  }, {
    id: "",
    AlphaNumber: "QQ1111",
    terms: "3"
  }, {
    id: "ABC12346",
    AlphaNumber: "RR4444",
    terms: "3"
  }, {
    id: "",
    AlphaNumber: "SS1111",
    terms: "5"
  }, {
    id: "ABC12347",
    AlphaNumber: "ASQE223",
    terms: "5"
  }],

  obj = {};

for (const o of array) obj[o.terms]?.id || (obj[o.terms] = o);

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