Question

我在JSON中有一个像这样的列表：

var arr=[
{
     "text":"text1",
     "id"  :"1",
     //no parent id
 },
{
     "text":"text2",
     "id"  :"2",
     "idParent":"1"
 },
{
     "text":"text3",
     "id"  :"3",
     "idParent":"2"
 },
{
     "text":"text4",
     "id"  :"4",
     "idParent":"1"
 },
{
     "text":"text5",
     "id"  :"5",
      //no parent id
 },
];

我需要在层次结构树中转换该列表，如下所示：

var arr = [
  {
    "text": "Parent 1",
    "id"  : "1",
    "nodes": [
      {
        "text": "Child 1",
        "id"  : "2",
        "parentid"  : "1",
        "nodes": [
          {
            "text": "Grandchild 1",
            "id"  : "4",
            "parentid"  : "2",
          },
          {
            "text": "Grandchild 2",
             "id"  : "8",
            "parentid"  : "2",
          }
        ]
      },
      {
        "text": "Child 2",
        "id"  : "10",
        "parentid"  : "1",
      }
    ]
  },
  {
    "text": "Parent 2",
    "id"  : "19",
    //no parent id
  }
];

只有根父母没有“parentid”元素。

¿我怎样才能在Javascript中执行此操作？

提前感谢您的帮助。

Answer 1

无序节点的解决方案。

var arr = [
        { text: "text3", id: "3", parentId: "2" },
        { text: "text2", id: "2", parentId: "1" },
        { text: "text4", id: "4", parentId: "1" },
        { text: "text1", id: "1", /* no parent id */ },
        { text: "text5", id: "5", /* no parent id */ }
    ],
    data = arr.reduce(function (r, a) {
        function getParent(s, b) {
            return b.id === a.parentId ? b : (b.nodes && b.nodes.reduce(getParent, s));
        }

        var index = 0, node;
        if ('parentId' in a) {
            node = r.reduce(getParent, {});
        }
        if (node && Object.keys(node).length) {
            node.nodes = node.nodes || [];
            node.nodes.push(a);
        } else {
            while (index < r.length) {
                if (r[index].parentId === a.id) {
                    a.nodes = (a.nodes || []).concat(r.splice(index, 1));
                } else {
                    index++;
                }
            }
            r.push(a);
        }
        return r;
    }, []);

document.write('<pre>' + JSON.stringify(data, 0, 4) + '</pre>');

Answer 2

我在这里使用了一个对象作为地图，否则你必须遍历所有数组元素以搜索节点的id，并且删除元素也更容易。 toString（）函数在那里，因为map的键必须是字符串。

// we use a map for search for the nodes by their ids
var nodes_map = {};

// we copy all the elements into the map, where the keys are the ids of the nodes
for ( var i=0; i<arr.length; ++i )
{
    // create nodes array for each node
    arr[ i ].nodes = [];

    // copy into the map
    nodes_map[ arr[i].id.toSting() ] = arr[ i ];
}    

// we iterate through all nodes, and add them into their parent's node array
for ( var key in nodes_map )
{
    // current node
    var node = nodes_map[ key ];

    // if the current node have idParent property, and the parent exists in the map
    if ( "idParent" in node && node.idParent.toSting() in nodes_map )
    {
        // we add the current node to the parent's nodes array 
        nodes_map[ node.idParent.toSting() ].nodes.push( node );
    }
}

// we remove all the nodes from the map, that has parents
for ( var key in nodes_map )
{
    // current node
    var node = nodes_map[ key ];

    // if it has idParent property
    if ( "idParent" in node )
    {
        // we remove from the map
        delete nodes_map[ key ];
    }
}    

// results array
var new_arr = [];
// copy back the nodes from the map into an array
for ( var key in nodes_map )
{
    new_arr.push( nodes_map[ key ] );
}

如何从JSON数据创建树结构

2 个答案: