javascript - 从数据中为 jqtree 构建一个树 json

Question

我正在尝试为以下构建 jqtree 的数据创建树 json

var data = [
    {
        "director": "Name1",
        "name": "sub-sub-child1",
        "teamname": "sub-child1",
        "teamManager": "child1"
    },

    {
       "director": "Name1",
       "name": "sub-sub-child2",
       "teamname": "sub-child1",
       "teamManager": "child1"
    },
    {
        "director": "Name1",
        "name": "sub-sub-child2",
        "teamname": "sub-child2",
        "teamManager": "child2"
    },
    {
        "director": "Name2",
        "name": "sub-sub-child1",
        "teamname": "sub-child1",
        "teamManager": "child1"
    },
    {
        "director": "Name2",
        "name": "sub-sub-child2",
        "teamname": "sub-child2",
        "teamManager": "child2"
    },
    {
        "director": "Name3",
        "name": "sub-sub-child1",
        "teamname": "sub-child1",
        "teamManager": "child1"
    },
    {
        "director": "Name3",
        "name": "sub-sub-child2",
        "teamname": "sub-child2",
        "teamManager": "child2"
    }
];

console.log(data);

var aNode = [];
                var aTempNode  = [];
                $.each(data, function(index, value) {
                    //console.log(value);
                    var key = {};
                    if($.inArray(value.director, aTempNode) == -1) {
                        aTempNode.push(value.director);
                        key['label'] = value.director;
                        key['children'] = [{label: value.teamManager, children: [{label: value.teamname, children: [{label: value.name}]}]}];
                        aNode.push(key);
                    } else {
                        //console.log(aNode)
                        if(key['teamname'] == aNode.children) {

                        }
                    }

                });
                console.log(aNode);

小提琴

编辑2

树应该是这种形式

Name1
 |
 |___child1
 |     |
 |     |___sub-child1
 |     |      |
 |     |      |___sub-sub-child1
 |     |      |___sub-sub-child2
 |     |
 |     |___sub-child2
 |            |
 |            |___sub-sub-child1
 |            |___sub-sub-child2
 |
 |____child2  

编辑1

我已经完成了一半。我很困惑如何从这里开始。 小提琴

Answer 1

试试这个，你会在“temp”变量中得到你需要的东西

var temp = [];
$.each(data, function(row, val) {
	var director = $.grep(temp, function(v) {
		return v.label == val.director
	});

	if (director.length) {
		var teamManager = $.grep(director[0].children, function(v) {
			return v.label == val.teamManager
		})
		if (teamManager.length) {
			var teamname = $.grep(teamManager[0].children, function(v) {
				return v.label == val.teamname
			})
			if (teamname.length) {
				var name = $.grep(teamname[0].children, function(v) {
					return v.label == val.name
				})
				if (!name.length) {
					teamname[0].children.push({
						label: val.name,
						children: []
					});
				}
			} else {
				teamManager[0].children.push({
					label: val.teamname,
					children: [{
						label: val.name,
						children: []
					}]
				});
			}

		} else {
			director[0].children.push({
				label: val.teamManager,
				children: [{
					label: val.teamname,
					children: [{
						label: val.name,
						children: []
					}]
				}]
			})
		}
	} else {
		temp.push({
			label: val.director,
			children: [{
				label: val.teamManager,
				children: [{
					label: val.teamname,
					children: [{
						label: val.name,
						children: []
					}]
				}]
			}]
		});
	}

});