Question

我的密钥值对的JSON为：

      let countByView= {
          "count1": "0",
          "count2": "230",
          "count3": "246",
          "count4": "172",
          "view1": "2017",
          "view2": "2018",
          "view3": "2019",
          "view4": "2020"
        }

我希望转化为

 let countByView=
        [
          {
            "count1": 0,
            "view1": 2017
          },
          {
            "count1": 230,
            "view1": 2018
          },
          {
            "count1": 246,
            "view1": 2019
          },
          {
            "count1": 172,
            "view1": 2020
          }
        ]

我尝试使用下面的代码，但没有得到预期的结果，因为我无法正确转换。请让我知道有没有更好的解决方案..

  let result = Object.keys(countByView).map(e => {
        let ret = {};
        ret[e] = obj[e];
        return ret;
    });

console.log(result);

Answer 1

您想要的数组长度是countByView的键数的一半。使用长度，您可以生成带有for循环的有序数组或Array#from：

＆＃13;

const countByView = {"count1":"0","count2":"230","count3":"246","count4":"172","view1":"2017","view2":"2018","view3":"2019","view4":"2020"}

const length = Object.keys(countByView).length / 2

const result = Array.from({ length }, (_, i) => ({
  count: countByView[`count${i + 1}`],
  view: countByView[`view${i + 1}`]
}))

console.log(result)

＆＃13;

Answer 2

您可以对具有数字的相同命名对象使用哈希表。

var countByView = { count1: "0", count2: "230", count3: "246", count4: "172", view1: "2017", view2: "2018", view3: "2019", view4: "2020" },
    hash = Object.create(null),
    result = Object.keys(countByView).reduce(function (r, k) {
        var [ , key, value] = k.match(/^(\D+)(\d+)$/);
        if (!hash[value]) {
            r.push(hash[value] = {});
        }
        hash[value][key] = countByView[k];
        return r;
    }, []);

console.log(result);

Answer 3

您可以简单地遍历该json对象的Object.keys并添加一些逻辑来获取该结构。

＆＃13;

let countByView= {
    "count1": "0",
    "count2": "230",
    "count3": "246",
    "count4": "172",
    "view1": "2017",
    "view2": "2018",
    "view3": "2019",
    "view4": "2020"
  };
var keys = Object.keys(countByView);
var resArray = [];
for(var i=0; i<keys.length; i++){
  var key = keys[i];
  var obj = {};
  if(key.indexOf('count') !== -1){
    obj[key] = countByView[key];
    obj['view'+(i+1)] = countByView['view'+(i+1)]; 
    resArray.push(obj);
  }
}

console.log(resArray);

＆＃13;

Answer 4

您的示例包含以索引结尾的标签。您还没有指定它是否总是1,2 ......或者它可以是任何数字。在我的下面的代码中，我假设它可以是任何数字（只要 count 和 view 具有相同的数字）。

＆＃13;

let countByView= {
  "count1": "0",
  "count2": "230",
  "count3": "246",
  "count4": "172",
  "view1": "2017",
  "view2": "2018",
  "view3": "2019",
  "view4": "2020"
}

function transform(input) {
  return Object
    .keys(input)
    .filter( key => key.indexOf('count') === 0)
    .reduce(
    (list, key) => {
      // Using a regexp to split the key into name and index.
      let item = /^([a-z]+)([0-9]+)$/.exec(key);
      if (!item) return console.warn('illegal key', key);
      let index = item[2];
      // push the data to the list, also converts the value to a number.
      list.push({
        count : input['count' + index ],
        view : +input['view' + index ]
      })
      return list;
    },[]
  );
}

console.log(transform(countByView));

＆＃13;

将键值对转换为所需对象的数组

4 个答案: