如何將物件中的專案轉換為陣列串列

{
    Type: "Fire",
    Name: "Shark",
    StartRange0: "10",
    EndRange0: "20",
    StartRange1: "30",
    EndRange1: "40",
    StartRange2: "60",
    EndRange2: "70"
}

在上述物件中，StartRange 和 EndRange 可以是不同物件中的更多項。即這些值是從回圈中獲得的（StartRange index），所以可能有 StartRange1,2,3... lly, EndRange1,2,3...

上述物件結構必須更改為，

{
   "Fire":{
      "Name":"Shark",
      "List":[
         [
            "10",
            "20"
         ],
         [
            "30",
            "40"
         ],
         [
            "60",
            "70"
         ]
      ]
   }
}

物件內部的 List 陣列是 [StartRange0,EndRange0],[StartRange1,EndRange2]... 的組合... 以此類推...

我試過了，

let newObj = {}
newObj[oldObj.Type] = {
    Name: oldObj.Name, //oldObj is the first object defined above.
    List: [[oldObj.StartRange0,oldObj.EndRange0],[oldObj.tartRange1,oldObj.EndRange1],[oldObj.StartRange2,oldObj.EndRange2]]
}

但在這里，它僅限于 StartRange2 和 EndRange2，但我需要 StartRangeN 和 EndRangeN。

uj5u.com熱心網友回復：

您可以從物件中的所有鍵中提取數字后綴，對它們進行排序，然后構建匹配和值StartRange*的陣列。StartRangeEndRange

const oldObj = {
  Type: "Fire",
  Name: "Shark",
  StartRange0: "10",
  EndRange0: "20",
  StartRange1: "30",
  EndRange1: "40",
  StartRange2: "60",
  EndRange2: "70",
};

// Extract Type, Name and the rest
const { Type, Name, ...ranges } = oldObj;

// A regex to match "StartRange*" keys and extract suffixes
const startRangeRx = /^StartRange(\d )$/;

const newObj = {
  [Type]: {
    Name,
    List: Object.keys(ranges)
      .reduce((keys, k) => {
        // Extract suffixes as numbers
        const suffix = k.match(startRangeRx)?.[1];
        if (suffix !== undefined && `EndRange${suffix}` in ranges) {
          return [...keys, Number(suffix)];
        }
        return keys;
      }, [])
      .sort() // sort
      .map((suffix) => {
        return [ranges[`StartRange${suffix}`], ranges[`EndRange${suffix}`]];
      }),
  },
};

console.log(newObj);

.as-console-wrapper { max-height: 100% !important; }

uj5u.com熱心網友回復：

你可以玩這個例子

const object = {
  Type: "Fire",
  Name: "Shark",
  StartRange0: "10",
  EndRange0: "20",
  StartRange1: "30",
  EndRange1: "40",
  StartRange2: "60",
  EndRange2: "70"
};

let index = 0;
let isFinished = false;
const results = [];

while (!isFinished) {
  const startKey = `StartRange${index}`;
  const endKey = `EndRange${index}`;

  isFinished = !object.hasOwnProperty(startKey) || !object.hasOwnProperty(endKey);

  !isFinished && results.push([object[startKey], object[endKey]]);

  index  ;
}

console.log(results);

uj5u.com熱心網友回復：

這就是我做地圖的方式

const input = {
  Type: "Fire",
  Name: "Shark",
  StartRange0: "10",
  EndRange0: "20",
  StartRange1: "30",
  EndRange1: "40",
  StartRange2: "60",
  EndRange2: "70"
}

const toList = (l) => Array.from({
  length: Object.keys(l).length / 2
}, (_, N) => [l[`StartRange${N}`], l[`EndRange${N}`]])

const toOutput = ({ Type, Name, ...rest }) => {
  const List = toList(rest)
  return ({
    [Type]: {
      Name,
      List
    }
  })
}

const output = toOutput(input)
console.log(output)

uj5u.com熱心網友回復：

這可能有效，但如果您可以更改該結構并使用另一個結構會更好。

const test = {
  Type: "Fire",
  Name: "Shark",
  StartRange0: "10",
  EndRange0: "20",
  StartRange1: "30",
  EndRange1: "40",
  StartRange2: "60",
  EndRange2: "70",
};

const format = (obj) => {
  const { Type, Name, ...rest } = obj;
  const list = [];
  const keys = Object.keys(rest);
  for (let index = 0; index < keys.length; index  ) {
    const start = rest[`StartRange${index}`];
    const end = rest[`EndRange${index}`];
    if (start && end) {
      list.push([start, end]);
    } else {
      break; // assuming you don't have any other elements
    }
  }
  return {
    [Type]: {
      Name,
      List: list,
    },
  };
};

console.log(JSON.stringify(format(test)));

標籤：javascript 目的

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