在 javascript 中映射嵌套数组

Question

我有一个 objects 的嵌套数组，如下所示：

let data = [
  {
      id: 1,
      title: "Abc",
      children: [
          {
              id: 2,
              title: "Type 2",
              children: [
                  {
                      id: 23,
                      title: "Number 3",
                      children:[] /* This key needs to be deleted */
                  }
              ]
          },
      ]
  },
  {
      id: 167,
      title: "Cde",
      children:[] /* This key needs to be deleted */
  }
]

我想要的只是递归地找到没有 children 的 leaves（当前是一个空数组）并从中删除 children 属性。

这是我的代码：

normalizeData(data, arr = []) {
    return data.map((x) => {
        if (Array.isArray(x))
            return this.normalizeData(x, arr)
        return {
            ...x,
            title: x.name,
            children: x.children.length ? [...x.children] : null
        }
    })
}

Answer 1

你需要为此使用递归：

let data = [{
    id: 1,
    title: "Abc",
    children: [{
      id: 2,
      title: "Type 2",
      children: [{
        id: 23,
        title: "Number 3",
        children: [] /* This key needs to be deleted */
      }]
    }]
  },
  {
    id: 167,
    title: "Cde",
    children: [] /* This key needs to be deleted */
  }
]

function traverse(obj) {
  for (const k in obj) {
    if (typeof obj[k] == 'object' && obj[k] !== null) {
      if (k === 'children' && !obj[k].length) {
        delete obj[k]
      } else {
        traverse(obj[k])              
      }
    }
  }
}

traverse(data)
console.log(data)

Answer 2

Nik 的回答很好（尽管我看不出那样访问 children 键有什么意义），但如果有帮助，这里有一个更短的替代方法：

let data = [
  {id: 1, title: "Abc", children: [
    {id: 2, title: "Type 2", children: [
      {id: 23, title: "Number 3", children: []}
    ]}
  ]},
  {id: 167, title: "Cde", children: []}
];

data.forEach(deleteEmptyChildren = o => 
  o.children.length ? o.children.forEach(deleteEmptyChildren) : delete o.children);

console.log(data);

如果children不是一直都有，可以把代码的主要部分改成：

data.forEach(deleteEmptyChildren = o => 
  o.children && o.children.length 
    ? o.children.forEach(deleteEmptyChildren) 
    : delete o.children);

Answer 3

你可以使用递归来做到这一点。

所以这里的基本思想是在 removeEmptyChild 函数中我们检查 children 长度是否为非零。因此，如果是，我们遍历 children 数组中的每个元素并将它们作为参数再次传递给函数，如果 children 长度为零，我们删除 children 键。

let data=[{id:1,title:"Abc",children:[{id:2,title:"Type2",children:[{id:23,title:"Number3",children:[]}]},]},{id:167,title:"Cde",children:[]},{id:1}]

function removeEmptyChild(input){
  if( input.children && input.children.length ){
    input.children.forEach(e => removeEmptyChild(e) )
  } else {
    delete input.children
  }
  return input
}

data.forEach(e=> removeEmptyChild(e))

console.log(data)

Answer 4

只需使用 forEach 进行简单递归即可。

let data = [{
    id: 1,
    title: "Abc",
    children: [{
      id: 2,
      title: "Type 2",
      children: [{
        id: 23,
        title: "Number 3",
        children: [] /* This key needs to be deleted */
      }]
    }, ]
  },
  {
    id: 167,
    title: "Cde",
    children: [] /* This key needs to be deleted */
  }
]

const cleanUp = data =>
  data.forEach(n =>
    n.children.length
      ? cleanUp(n.children)
      : (delete n.children))
      
      
cleanUp(data)
console.log(data)

这假设 children 在那里。如果它可能丢失，则只需要对检查进行微小更改，这样它就不会在长度检查中出错。 n.children && n.children.length

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