5

我有一组与此类似的深层 JSON 对象：

var hierarchy = [
  {
    "title": "category 1",
    "children": [
      {"title": "subcategory 1",
        "children": [
          {"id": 1, "title": "name 1"},
          {"id": 2, "title": "name 2"},
          {"id": 3, "title": "name 3"}
        ]
      },
      {"title": "subcategory 2",
        "children": [
          {"id": 1, "title": "name 4"}
        ]
      }
    ]
  },
  {
    "title": "category 2",
    "children": [etc. - shortened for brevity]
  }
];

所以基本上它是一个层次结构 - 有些类别可以有子类别，其中包含具有一些 ID 和名称的对象。我还有一组与最深层次结构级别（没有子对象）相关的 ID，我需要过滤这组对象，以便只保留包含已定义对象的（子）类别。

例如，如果我有一个包含两个 ID 的数组：

var IDs = [2, 3];

结果将是：

var hierarchy = [
  {
    "title": "category 1",
    "children": [
      {"title": "subcategory 1",
        "children": [
          {"id": 2, "title": "name 2"},
          {"id": 3, "title": "name 3"}
        ]
      }
    ]
  }
];

即整个，整个“类别2”对象被移除，整个“子类别2”被移除，ID为“1”的对象被移除。

问题是这些对象的深度是可变且未知的 - 有些对象没有孩子，有些对象有孩子也有孩子等等，任何子类别本身都可以有一个子类别，我基本上需要找到没有孩子的对象定义 ID 并保留每个 ID 的完整路径。

谢谢你。

4

4 回答 4

6

基本上，对树执行深度优先遍历，在每个节点上调用回调函数。如果该节点是叶节点并且它的 ID 出现在您的列表中，则克隆通向该叶的分支，但不要重新克隆已克隆的分支的任何部分。

构建树的部分和过滤副本后，您需要清理原始副本。为了记账的目的，我在这个过程中对原始树进行了变异——跟踪哪些分支已经被克隆。

编辑：修改代码以过滤树列表，而不仅仅是一棵树

var currentPath = [];

function depthFirstTraversal(o, fn) {
    currentPath.push(o);
    if(o.children) {
        for(var i = 0, len = o.children.length; i < len; i++) {
            depthFirstTraversal(o.children[i], fn);
        }
    }
    fn.call(null, o, currentPath);
    currentPath.pop();
}

function shallowCopy(o) {
    var result = {};
    for(var k in o) {
        if(o.hasOwnProperty(k)) {
            result[k] = o[k];
        }
    }
    return result;
}

function copyNode(node) {
    var n = shallowCopy(node);
    if(n.children) { n.children = []; }
    return n;
}

function filterTree(root, ids) {
    root.copied = copyNode(root); // create a copy of root
    var filteredResult = root.copied;

    depthFirstTraversal(root, function(node, branch) {
        // if this is a leaf node _and_ we are looking for its ID
        if( !node.children && ids.indexOf(node.id) !== -1 ) {
            // use the path that the depthFirstTraversal hands us that
            // leads to this leaf.  copy any part of this branch that
            // hasn't been copied, at minimum that will be this leaf
            for(var i = 0, len = branch.length; i < len; i++) {
                if(branch[i].copied) { continue; } // already copied

                branch[i].copied = copyNode(branch[i]);
                // now attach the copy to the new 'parellel' tree we are building
                branch[i-1].copied.children.push(branch[i].copied);
            }
        }
    });

    depthFirstTraversal(root, function(node, branch) {
        delete node.copied; // cleanup the mutation of the original tree
    });

    return filteredResult;
}

function filterTreeList(list, ids) {
    var filteredList = [];
    for(var i = 0, len = list.length; i < len; i++) {
        filteredList.push( filterTree(list[i], ids) );
    }
    return filteredList;
}

var hierarchy = [ /* your data here */ ];
var ids = [1,3];

var filtered = filterTreeList(hierarchy, ids);
于 2013-10-20T23:07:03.427 回答
1

您可以使用扩展中的filterDeep方法：deepdashlodash

var obj = [{/* get Vijay Jagdale's source object as example */}];
var idList = [2, 3];
var found = _.filterDeep(
  obj,
  function(value) {
    return _.indexOf(idList, value.id) !== -1;
  },
  { tree: true }
);

filtrate对象将是：

[ { title: 'category 1',
    children:
     [ { title: 'subcategory 11',
         children:
          [ { id: 2, title: 'name 2' },
            { id: 3, title: 'name 3' } ] } ] },
  { title: 'category 2',
    children:
     [ { title: 'subcategory 21',
         children: [ { id: 3, title: 'name cat2sub1id3' } ] } ] } ]

这是您的用例的完整工作测试

于 2019-01-15T18:01:27.743 回答
0

虽然这是一个老问题，但我会加我的 2 美分。该解决方案需要通过循环、子循环等进行直接迭代，然后比较 ID 并构建结果对象。我有纯 JavaScript 和 jQuery 解决方案。虽然纯 javascript 适用于上面的示例，但我会推荐 jQuery 解决方案，因为它更通用，并且对对象进行“深度复制”，以防万一您有大型和复杂的对象，您不会遇到错误。

function jsFilter(idList){
  var rsltHierarchy=[];
  for (var i=0;i<hierarchy.length;i++) {
    var currCatg=hierarchy[i];
    var filtCatg={"title":currCatg.title, "children":[]};
    for (var j=0;j<currCatg.children.length;j++) {
  	var currSub=currCatg.children[j];
  	var filtSub={"title":currSub.title, "children":[]}
  	for(var k=0; k<currSub.children.length;k++){
  		if(idList.indexOf(currSub.children[k].id)!==-1)
  		   filtSub.children.push({"id":currSub.children[k].id, "title":currSub.children[k].title});
  	}
  	if(filtSub.children.length>0)
  		filtCatg.children.push(filtSub);
    }
    if(filtCatg.children.length>0)
  	rsltHierarchy.push(filtCatg);
  }
  return rsltHierarchy;
}

function jqFilter(idList){
  var rsltHierarchy=[];
  $.each(hierarchy, function(index,currCatg){
      var filtCatg=$.extend(true, {}, currCatg);
      filtCatg.children=[];
  	$.each(currCatg.children, function(index,currSub){
        var filtSub=$.extend(true, {}, currSub);
  	  filtSub.children=[];
  	  $.each(currSub.children, function(index,currSubChild){
  		if(idList.indexOf(currSubChild.id)!==-1)
  		  filtSub.children.push($.extend(true, {}, currSubChild));
        });
  	  if(filtSub.children.length>0)
  		filtCatg.children.push(filtSub);
      });
      if(filtCatg.children.length>0)
  	  rsltHierarchy.push(filtCatg);
  });
  return rsltHierarchy;
}

//Now test the functions...
var hierarchy = eval("("+document.getElementById("inp").value+")");
var IDs = eval("("+document.getElementById("txtBoxIds").value+")");

document.getElementById("oupJS").value=JSON.stringify(jsFilter(IDs));
$(function() {
   $("#oupJQ").text(JSON.stringify(jqFilter(IDs)));
});
#inp,#oupJS,#oupJQ {width:400px;height:100px;display:block;clear:all}
#inp{height:200px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

ID List: <Input id="txtBoxIds" type="text" value="[2, 3]">

<p>Input:
<textarea id="inp">[
  {
    "title": "category 1",
    "children": [
      {"title": "subcategory 11",
        "children": [
          {"id": 1, "title": "name 1"},
          {"id": 2, "title": "name 2"},
          {"id": 3, "title": "name 3"}
        ]
      },
      {"title": "subcategory 12",
        "children": [
          {"id": 1, "title": "name 4"}
        ]
      }
    ]
  },
  {
    "title": "category 2",
    "children": [
      {"title": "subcategory 21",
        "children": [
          {"id": 3, "title": "name cat2sub1id3"},
          {"id": 5, "title": "name cat2sub1id5"}
        ]
      },
      {"title": "subcategory 22",
        "children": [
          {"id": 6, "title": "name cat2sub2id6"},
          {"id": 7, "title": "name cat2sub2id7"}
        ]
      }
    ]
  }
]</textarea>

<p>Pure-Javascript solution results:
<textarea id="oupJS"></textarea>

<p>jQuery solution results:
<textarea id="oupJQ"></textarea>

于 2015-11-23T21:34:30.190 回答
0

我不会重新发明轮子。我们现在对大部分数据处理使用对象扫描，它很好地解决了您的问题。这是如何

// const objectScan = require('object-scan');

const filter = (input, ids) => {
  objectScan(['**[*]'], {
    filterFn: ({ value, parent, property }) => {
      if (
        ('id' in value && !ids.includes(value.id))
        || ('children' in value && value.children.length === 0)
      ) {
        parent.splice(property, 1);
      }
    }
  })(input);
};

const hierarchy = [ { title: 'category 1', children: [ { title: 'subcategory 1', children: [ { id: 1, title: 'name 1' }, { id: 2, title: 'name 2' }, { id: 3, title: 'name 3' } ] }, { title: 'subcategory 2', children: [ { id: 1, title: 'name 4' } ] } ] }, { title: 'category 2', children: [] } ];

filter(hierarchy, [2, 3]);

console.log(hierarchy);
// => [ { title: 'category 1', children: [ { title: 'subcategory 1', children: [ { id: 2, title: 'name 2' }, { id: 3, title: 'name 3' } ] } ] } ]
.as-console-wrapper {max-height: 100% !important; top: 0}
<script src="https://bundle.run/object-scan@13.8.0"></script>

免责声明：我是对象扫描的作者

于 2020-11-18T05:49:57.230 回答