遍歷javascript中的縮進陣列-有解無憂

我有一個 javascript 物件，如下所示：

let hogwartsHeirarchy = {
  Headmaster: [
    {
      name: "Professor Dumbledore",
      key: 1,
      Headmistress: [
        {
          name: "Minerva McGonagall",
          key: 2,
          StandByProfessor: [
            {
              name: "Rubeus Hagrid",
              subject: "Potions Master",
              key: 3,
              Professor: [
                { name: "Horace Slughorn", key: 4 },
                { name: "Severus Snape", key: 4 },
              ],
            },
            {
              name: "Alastor Moody",
              subject: "Defense Against the Dark Arts",
              key: 3,
              Professor: [
                { name: "Remus Lupin", key: 4 },
                { name: "Gilderoy Lockhart", key: 4 },
              ],
            },
          ],
        },
      ],
    },
  ],
};

我想列印/獲取每個節點值 [headmaster, headmastress,..] 及其相應的子值。我嘗試了各種方法，例如使用 for 回圈、遞回等回圈遍歷陣列，但不幸的是我無法從節點中獲取任何值。請幫忙。

例如：我用過這個：

printArray(hogwartsHeirarchy);

function printArray(arr){
    for(var i = 0; i < arr.length; i  ){
        if(arr[i] instanceof Array){
            console.log("true: ");

            console.log("intermediate one : ",arr[i]);

            printArray(arr[i]);

        }else{
            console.log("final one : ",arr[i]);
        }
    }
}

這些值可以以這種格式顯示：

Headmaster - name : Professor Dumbledore, key : 1
.
.
StandByProfessor - name : Robeus Hagrid, subject : Potion Master, key : 3
StandByProfessor - name : Alastor Moody, subject : Defence against the dark arts, key : 3
.
.
Professor - ...
Professor - ...
Professor - ...
Professor - ...

uj5u.com熱心網友回復：

我建議重組，以便始終使用相同的密鑰訪問下屬，從而可以非常簡單地訪問。我也做了它，所以每個節點都是一個人，頂部沒有不是人的物體。我留下了變數名，但它現在直接指代鄧布利多。

let hogwartsHeirarchy =
  {
    name: "Professor Dumbledore",
    role: "Headmaster",
    key: 1,
    subordinates: [
      {
        name: "Minerva McGonagall",
        role: "Headmistress",
        key: 2,
        subordinates: [
          {
            name: "Rubeus Hagrid",
            role: "StandByProfessor",
            subject: "Potions Master",
            key: 3,
            subordinates: [
              { name: "Horace Slughorn", key: 4, role: "Professor" },
              { name: "Severus Snape", key: 4, role: "Professor"  },
            ],
          },
          {
            name: "Alastor Moody",
            role: "StandByProfessor",
            subject: "Defense Against the Dark Arts",
            key: 3,
            subordinates: [
              { name: "Remus Lupin", key: 4, role: "Professor" },
              { name: "Gilderoy Lockhart", key: 4, role: "Professor" },
            ],
          },
        ],
      },
    ],
  };
function visitStaff(staffMember) {
    if (staffMember.subordinates) {
        for (const subordinate of staffMember.subordinates) {
            visitStaff(subordinate);
        }
    }
    console.log("Staff member:", staffMember);
}
visitStaff(hogwartsHeirarchy);

在設定資料結構時，重要的是要考慮如何訪問它，以及它的定義部分是什么。在這種情況下，人是節點，（從屬）關系是圖的邊。

在你的原始代碼中，你有一個物件{ Headmaster: [...] }——它代表什么？它是一個人嗎？不; 是關系嗎？有點，但沒有。它定義了一些關于Dumbledoor，他是校長，但不是誰或什么他是校長的。所以它實際上只是描述了鄧布利多的角色/職位，所以它作為一個人的財產更有意義。它作為一個物件是多余的。

它有助于對齊您的物件，以便它們都代表某些東西。您應該能夠描述每個物件和陣列是什么。

uj5u.com熱心網友回復：

您可以從物件中洗掉已知鍵并獲取型別層次結構，然后迭代屬性并僅在型別存在時回傳型別、名稱、主題和鍵的元組。

const
    getValues = (object, type) => [
        ...(type ? [[type, object.name, object.subject || '', object.key]] : []),
        ...Object
            .keys(object)
            .filter(k => !['name', 'subject', 'key'].includes(k))
            .flatMap(k => object[k].flatMap(o => getValues(o, k)))
        ],
    hogwartsHierarchy = { Headmaster: [{ name: "Professor Dumbledore", key: 1, Headmistress: [{ name: "Minerva McGonagall", key: 2, StandByProfessor: [{ name: "Rubeus Hagrid", subject: "Potions Master", key: 3, Professor: [{ name: "Horace Slughorn", key: 4 }, { name: "Severus Snape", key: 4 }] }, { name: "Alastor Moody", subject: "Defense Against the Dark Arts", key: 3, Professor: [{ name: "Remus Lupin", key: 4 }, { name: "Gilderoy Lockhart", key: 4 }] }] }] }] };

console.log(getValues(hogwartsHierarchy));

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uj5u.com熱心網友回復：

給定資料結構：

a) 我假設每種型別的“標題”中只有一個陣列，并且
b) 該陣列將包含一個與其父物件結構相似的物件串列

這是可能的 ...

使用for..of以
遍歷物件上的每個鍵，并將它們添加到字串中。因為有包含物件的陣列，
我可以遍歷它們，然后
做一個遞回回圈。

const hogwartsHierarchy = { Headmaster: [{ name: "Professor Dumbledore", key: 1, Headmistress: [{ name: "Minerva McGonagall", key: 2, StandByProfessor: [{ name: "Rubeus Hagrid", subject: "Potions Master", key: 3, Professor: [{ name: "Horace Slughorn", key: 4 }, { name: "Severus Snape", key: 4 }] }, { name: "Alastor Moody", subject: "Defense Against the Dark Arts", key: 3, Professor: [{ name: "Remus Lupin", key: 4 }, { name: "Gilderoy Lockhart", key: 4 }] }] }] }] };

function printAllWithChilds(obj, prevProp) {
  let listItem = (prevProp) ? ' -- '   prevProp : '';
  
  for (const property in obj) {    // 1
    if (obj[property] instanceof Array) {
       obj[property].forEach((child_obj) => {                   // 3
         listItem  = printAllWithChilds(child_obj, property);   // 4
       });
    } else {
      listItem  = `, ${property}: ${obj[property]}`;            // 2
    }
  }
  
  return listItem;
}

let listStr = printAllWithChilds(hogwartsHierarchy);
console.log(listStr);

老實說，我會hogwartsHierarchy按照一種資料庫結構分成更小的部分，primary_key每個人的位置都是獨一無二的。這些陣列沒有多大意義，直到您查看變數professors以及它們各自的belongs_to鍵如何對應于standbyprofessors，您可以看到“Horace Slughorn”屬于“Rubeus Hagrid”。

const headermaster = {
  name: "Professor Dumbledore",
  primary_key: 1
};

const headmistress = {
  name: "Minerva McGonagall",
  primary_key: 2,
  belongs_to: 1
};

const standbyprofessors = [{
    name: "Rubeus Hagrid",
    subject: "Potions Master",
    primary_key: 3,
    belongs_to: 2
  },
  {
    name: "Alastor Moody",
    subject: "Defense Against the Dark Arts",
    primary_key: 4,
    belongs_to: 2
  }
];

const professors = [{
    name: "Horace Slughorn",
    primary_key: 5,
    belongs_to: 3
  },
  {
    name: "Severus Snape",
    primary_key: 6,
    belongs_to: 3
  },
  {
    name: "Remus Lupin",
    primary_key: 7,
    belongs_to: 4
  },
  {
    name: "Gilderoy Lockhart",
    primary_key: 8,
    belongs_to: 4
  },
];

uj5u.com熱心網友回復：

基于1j01物件模型：

const hogwartsHeirarchy =
{  name: "Professor Dumbledore", role: "Headmaster", key: 1,
    subordinates: [{ name: "Minerva McGonagall", role: "Headmistress", key: 2,
        subordinates: [
          { name: "Rubeus Hagrid", role: "StandByProfessor", subject: "Potions Master", key: 3,
            subordinates: [
              { name: "Horace Slughorn", key: 4, role: "Professor" },
              { name: "Severus Snape", key: 4, role: "Professor"  }]},
          { name: "Alastor Moody", role: "StandByProfessor", subject: "Defense Against the Dark Arts", key: 3,
            subordinates: [
              { name: "Remus Lupin", key: 4, role: "Professor" },
              { name: "Gilderoy Lockhart", key: 4, role: "Professor" }]}]}]};
  
  
const visitStaff = (staffMember) => {
  const iter = (obj) => {
    const { name, role, key, subordinates } = obj;
    const subject = obj.subject ? `, subject: ${obj.subject}` : '';
    const line = `${role} - name : ${name}${subject}, key : ${key}`;
    const sublines = (Array.isArray(subordinates)) ? subordinates.flatMap((s) => iter(s)) : [];
    return [line, ...sublines];
  }
  return iter(staffMember).join('\n');
}
console.log(visitStaff(hogwartsHeirarchy));

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標籤：javascript 数组循环递归嵌套

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