Question

我目前有一个对象数组，我试图将ID作为对象键重新整形为嵌套对象，并使用ID定位parent id如果不是"0"。我已经尝试了几种方法，但是我很难过得很开心！对我来说，的主要障碍是超过一层或两层深的东西。理想情况下，我需要它是动态的，所以它可以处理任何深度，虽然在野外我怀疑它将超过3层。

这是传入的对象数组：

[ { ID: "1671", parent: "0", }, { ID: "1223", parent: "0", }, { ID: "1668", parent: "0", }, { ID: "1688", parent: "0", }, { ID: "1669", parent: "0", }, { ID: "1681", parent: "1669", }, { ID: "1680", parent: "1669", }, { ID: "1670", parent: "1669", }, { ID: "1682", parent: "1669", }, { ID: "1433", parent: "1682", }, { ID: "1684", parent: "1682", }, { ID: "1672", parent: "1684", }, { ID: "1685", parent: "1672", }, { ID: "1686", parent: "1672", }, { ID: "1683", parent: "0", }, { ID: "1230", parent: "0", }, { ID: "1667", parent: "0", }, { ID: "1687", parent: "0", } ];

这是理想的转型：

1671: { ID: "1671", parent: "0", }, 1223: { ID: "1223", parent: "0", }, 1668: { ID: "1668", parent: "0", }, 1688: { ID: "1688", parent: "0", }, 1669: { ID: "1669", parent: "0", children: { 1681: { ID: "1681", parent: "1669", }, 1680: { ID: "1680", parent: "1669", }, 1670: { ID: "1670", parent: "1669", }, 1682: { ID: "1682", parent: "1669", children: { 1433: { ID: "1433", parent: "1682", }, 1684: { ID: "1684", parent: "1682", children: { 1672: { ID: "1672", parent: "1684", children: { 1685: { ID: "1685", parent: "1672", }, 1686: { ID: "1686", parent: "1672", } } } } } } } } }, 1683: { ID: "1683", parent: "0", }, 1230: { ID: "1230", parent: "0", }, 1667: { ID: "1667", parent: "0", }, 1687: { ID: "1687", parent: "0", }

我的大多数失败的方法一直在使用Array.reduce()和其他类型的递归但无济于事，所以我有兴趣看看是否有解决方案。

由于

Answer 1

您可以使用这样的递归函数：

function findFor(parentId) {
  // create a new object to store the result
  var z = {};

  // for each item in a
  for (var i = 0; i<a.length; i++){

    // find all children of parentId
    if (a[i].parent === parentId) {

      // recursively find children for each children of parentId
      var ch = findFor(a[i].ID);

      // if it has no children, skip adding the children prop
      var o = Object.keys(ch).length === 0 ? {} : { children: ch };
      z[a[i].ID] = Object.assign(o, a[i]);
    }
  }

  return z;
}

// find all item whose parent is "0"
console.log(findFor("0"));

＆＃13;

var a = [
  { ID: "1671", parent: "0", },
  { ID: "1223", parent: "0", },
  { ID: "1668", parent: "0", },
  { ID: "1688", parent: "0", },
  { ID: "1669", parent: "0", },
  { ID: "1681", parent: "1669", },
  { ID: "1680", parent: "1669", },
  { ID: "1670", parent: "1669", },
  { ID: "1682", parent: "1669", },
  { ID: "1433", parent: "1682", },
  { ID: "1684", parent: "1682", },
  { ID: "1672", parent: "1684", },
  { ID: "1685", parent: "1672", },
  { ID: "1686", parent: "1672", },
  { ID: "1683", parent: "0", },
  { ID: "1230", parent: "0", },
  { ID: "1667", parent: "0", },
  { ID: "1687", parent: "0", }
];

function findFor(parentId) {
  var z = {};
  for (var i = 0; i<a.length; i++){
    if (a[i].parent === parentId) {
      var ch = findFor(a[i].ID);
      var o = Object.keys(ch).length === 0 ? {} : { children: ch };
      z[a[i].ID] = Object.assign(o, a[i]);
    }
  }
  
  return z;
}

console.log(findFor("0"));

＆＃13;

Answer 2

通过尊重父母的单循环非递归方法。

var flat = [{ ID: "1671", parent: "0" }, { ID: "1223", parent: "0" }, { ID: "1668", parent: "0" }, { ID: "1688", parent: "0" }, { ID: "1669", parent: "0" }, { ID: "1681", parent: "1669" }, { ID: "1680", parent: "1669" }, { ID: "1670", parent: "1669" }, { ID: "1682", parent: "1669" }, { ID: "1433", parent: "1682" }, { ID: "1684", parent: "1682" }, { ID: "1672", parent: "1684" }, { ID: "1685", parent: "1672" }, { ID: "1686", parent: "1672" }, { ID: "1683", parent: "0" }, { ID: "1230", parent: "0" }, { ID: "1667", parent: "0" }, { ID: "1687", parent: "0" }],
    tree = function (data, root) {
        var o = {};
        data.forEach(function(a) {
            a = Object.assign({}, a);
            if (o[a.ID] && o[a.ID].children) {
                a.children = o[a.ID].children;
            }
            o[a.ID] = { [a.ID]: a };
            o[a.parent] = o[a.parent] || {};
            o[a.parent][a.parent] = o[a.parent][a.parent] || {};
            o[a.parent][a.parent].children = o[a.parent][a.parent].children || {};
            o[a.parent][a.parent].children[a.ID] = a;
        });
        return o[root][root].children;
    }(flat, '0');

console.log(tree);

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从平面数组创建嵌套对象

2 个答案: