如何在javascript中為此回圈遍歷嵌套的物件陣列？

const array = [
{ 
id: "1",
children: [],
messages:[1, 'Text'],
entry: "Father",
},

{
id: "2",
entry: "Mother",
children: [
    {entry: "John Jr"},
    {entry: "Steven Jr"},
    {entry: "Tim Jr"},
    ],
messages:[2, 'Text'],
}, 

{
id: "3",
entry: "Son",
children: [
    {entry: "XXX Jr"},
    {entry: "Steven Jr"},
    {entry: "Tim Jr"}
    ],
 messages:[3, 'Text'],
 },
]

所以我想要的是，當我搜索“蒂姆”時。我希望輸出是

案例 1：當我搜索 Tim 時，我應該得到 id: 2 和 id: 3 的物件，因為 Tim 是兩個物件中都匹配的條目。所以我想以相同的順序回傳整個 id: 2 和 3 物件。

{
id: "2",
entry: "Mother",
children: [
    {entry: "John Jr"},
    {entry: "Steven Jr"},
    {entry: "Tim Jr"},
    ],
messages:[2, 'Text'],
}, 

{
id: "3",
entry: "Son",
children: [
    {entry: "XXX Jr"},
    {entry: "Steven Jr"},
    {entry: "Tim Jr"}
    ],
 messages:[3, 'Text'],
 }

案例 2：當我搜索 John 時，我應該得到第二個物件

{
id: "2",
entry: "Mother",
children: [
    {entry: "John Jr"},
    ],
messages:[2, 'Text'],
}

案例 3：當我搜索 Steven 時，我應該得到 id:2 和 id:3 作為

id: "2",
entry: "Mother",
children: [
    {entry: "John Jr"},
    {entry: "Steven Jr"},
    ],
messages:[2, 'Text'],
}, 

{
id: "3",
entry: "Son",
children: [
    {entry: "XXX Jr"},
    {entry: "Steven Jr"},
    ],
 messages:[3, 'Text'],
 }

我使用 .filter() 來過濾頂級，但這不是檢查孩子。

array.filter((item) => item.entry.toUpperCase().includes(text.toUpperCase())); 

任何幫助將不勝感激，因為我在這里完全迷失了 forEach。提前致謝

uj5u.com熱心網友回復：

您可以使用Array.Filter， Array.some并將子陣列與搜索entry物件的索引拼接

const array = [{
    id: "1",
    children: [],
    messages: [1, 'Text'],
    entry: "Father",
  },

  {
    id: "2",
    entry: "Mother",
    children: [{
        entry: "John Jr"
      },
      {
        entry: "Steven Jr"
      },
      {
        entry: "Tim Jr"
      },
    ],
    messages: [2, 'Text'],
  },

  {
    id: "3",
    entry: "Son",
    children: [{
        entry: "XXX Jr"
      },
      {
        entry: "Steven Jr"
      },
      {
        entry: "Tim Jr"
      }
    ],
    messages: [3, 'Text'],
  },
];


let searchText = 'John';

let result = array.filter(el => {
  let i;
  let found = el.children.some((e, j) => {
    i = j;
    return e.entry.toLowerCase().indexOf(searchText.toLowerCase()) >= 0;
  });

  if (found)
    el.children = el.children.splice(0, i   1);

  return found;
});

console.log(result);

uj5u.com熱心網友回復：

const data=[{id:"1",children:[],messages:[1,"Text"],entry:"Father"},{id:"2",entry:"Mother",children:[{entry:"John Jr", children: [{entry: 'Bob', children: [{entry: 'Marvin'},{entry: 'Summer', children: [{entry: 'Batman'}, {entry: 'Robin'}]}]},{entry: 'Dave', children: [{entry: 'Sue'}, {entry: 'Sam'}]}]},{entry:"Steven Jr"},{entry:"Tim Jr"}],messages:[2,"Text"]},{id:"3",entry:"Son",children:[{entry:"XXX Jr", children: [{entry: 'Batman'}]},{entry:"Steven Jr"},{entry:"Tim Jr"}],messages:[3,"Text"]}];

function recurseChildren(children) {
  const arr = [];
  function loop(children) {
    for (const entry of children) {
      arr.push(entry.entry);
      if (entry.children) {
        loop(entry.children);
      }
    }
  }
  loop(children);
  return arr;
}

function finder(data, name) {

  // `filter` over the array
  return data.filter(obj => {

    // Recurse over the children array of all entries
    // to compile a list of names
    const names = recurseChildren(obj.children);

    // For each object check to see if
    // at least one of the entries in children
    // includes the name
    return names.some(entry => {
      return entry.includes(name);
    });

  });
}

console.log(finder(data, 'Tim'));
console.log(finder(data, 'John'));
console.log(finder(data, 'Steven'));
console.log(finder(data, 'Batman'));

標籤：javascript 数组 目的

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