小白级别js数据转换问题，在线等

  • Z时代
  • 2024-02-23
  • 分类：IT

  col: [
        { class1: "增量效益", class2: "主营收入份额1", indexName: "未开始" },
        { class1: "增量效益", class2: "主营收入份额1", indexName: "已完成" },
        { class1: "增量效益", class2: "主营收入份额2", indexName: "未开始" },
        { class1: "增量效益", class2: "主营收入份额2", indexName: "已完成" },
        { class1: "网络效能", class2: "单基站流量1", indexName: "未采取" },
        { class1: "网络效能", class2: "单基站流量1", indexName: "已采取" },
        { class1: "网络效能", class2: "单基站流量2", indexName: "未采取" },
        { class1: "网络效能", class2: "单基站流量2", indexName: "已采取" },
      ],

以上数据格式的数据如何转成以下数据

formattArr: [
        {
          title: "增量效益",
          children: [
            {
              lable: "主营收入份额1",
              children: [{ lable: "未开始" }, { lable: "已完成" }],
            },
            {
              lable: "主营收入份额2",
              children: [{ lable: "未开始" }, { lable: "已完成" }],
            },
          ],
        },
        {
          title: "网络效能",
          children: [
            {
              lable: "单基站流量1",
              children: [{ lable: "未采取" }, { lable: "已采取" }],
            },
            {
              lable: "单基站流量2",
              children: [{ lable: "未采取" }, { lable: "已采取" }],
            },
          ],
        },
      ],

回答：

js">var col = [
    { class1: "增量效益", class2: "主营收入份额1", indexName: "未开始" },
    { class1: "增量效益", class2: "主营收入份额1", indexName: "已完成" },
    { class1: "增量效益", class2: "主营收入份额2", indexName: "未开始" },
    { class1: "增量效益", class2: "主营收入份额2", indexName: "已完成" },
    { class1: "网络效能", class2: "单基站流量1", indexName: "未采取" },
    { class1: "网络效能", class2: "单基站流量1", indexName: "已采取" },
    { class1: "网络效能", class2: "单基站流量2", indexName: "未采取" },
    { class1: "网络效能", class2: "单基站流量2", indexName: "已采取" },
];
function format(list) {
    var ret = [];
    for (var i = 0; i < list.length; ++i) {
        var item = list[i];
        var keys = [];
        for (var key in item) keys.unshift(key);
        for (var j = keys.length, k = 0, _ret = ret; j--; ++k) {
            var key = k ? "lable" : "title";
            var name = item[keys[j]];
            var obj = null;
            for (var l = 0; l < _ret.length; ++l) {
                if (_ret[l][key] === name) {
                    obj = _ret[l];
                    break;
                }
            }
            if (!obj) {
                obj = {};
                obj[key] = name;
                if (j) obj.children = [];
                _ret.push(obj);
            }
            if (obj.children) _ret = obj.children;
        }
    }
    return ret;
}
console.dir(format(col));

回答：

小白级别js数据转换问题，在线等

let obj ={ col: [
        { class1: "增量效益", class2: "主营收入份额1", indexName: "未开始" },
        { class1: "增量效益", class2: "主营收入份额1", indexName: "已完成" },
        { class1: "增量效益", class2: "主营收入份额2", indexName: "未开始" },
        { class1: "增量效益", class2: "主营收入份额2", indexName: "已完成" },
        { class1: "网络效能", class2: "单基站流量1", indexName: "未采取" },
        { class1: "网络效能", class2: "单基站流量1", indexName: "已采取" },
        { class1: "网络效能", class2: "单基站流量2", indexName: "未采取" },
        { class1: "网络效能", class2: "单基站流量2", indexName: "已采取" },
      ]}
let objMap = obj.col.reduce((acc,{class1,class2,indexName})=>((((acc[class1]||={})[class2])||=[]).push(indexName),acc),{});
console.log(objMap);
Object.entries(objMap).map(([title,titleChildren])=>({title,children:Object.entries(titleChildren).map(([lable,lableChildren])=>({lable,children:lableChildren.map(lable=>({lable}))}))}));

第一步先构造最简树形结构。

{"增量效益":{"主营收入份额1":["未开始","已完成"],"主营收入份额2":["未开始","已完成"]},"网络效能":{"单基站流量1":["未采取","已采取"],"单基站流量2":["未采取","已采取"]}}

存放成{}方便判断是否重复，如果直接构造数组[]的话，还需要遍历对比查找。
所以需要中间结构来转换。
第二步，最后就是将最简树形转换成有指定属性的树形数组。

[{"title":"增量效益","children":[{"lable":"主营收入份额1","children":[{"lable":"未开始"},{"lable":"已完成"}]},{"lable":"主营收入份额2","children":[{"lable":"未开始"},{"lable":"已完成"}]}]},{"title":"网络效能","children":[{"lable":"单基站流量1","children":[{"lable":"未采取"},{"lable":"已采取"}]},{"lable":"单基站流量2","children":[{"lable":"未采取"},{"lable":"已采取"}]}]}]

回答：

不在回答这种问题

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