如何在jquery jstree中获取已检查的节点

Question

我创建了一个jquery jstree,它工作正常.现在问题是如何获取已检查节点的详细信息.

用于创建JStree代码是:

$(function () {
$("#tree").jstree({ 
    "json_data" : {
        "data" : [
            {"data":"pe_opensourcescanning","id":0,"pId":-1,"children":  [{"data":"tags","id":30,"pid":0},{"data":"branches","id":29,"pid":0},{"data":"trunk","id":1,"pid":0,"children":[{"data":"import-export","id":28,"pid":1},{"data":"custom_development","id":12,"pid":1},{"data":"Connectors","id":7,"pid":1},{"data":"support","id":6,"pid":1},{"data":"Installation-Configuration","id":5,"pid":1},{"data":"backup","id":2,"pid":1}]}]}
        ]
    },
    "plugins" : [ "themes", "json_data", "checkbox", "ui" ]
}).bind("select_node.jstree", function (e, data) { alert(data.rslt.obj.data("id")); });

现在,在获取已检查的节点时,我需要这些已检查元素的所有属性值.比如"标签",json对象看起来像{"data":"tags","id":30,"pid":0},所以如果用户选择标签我需要"data"和"id"的值.我试图写一些代码,但不幸的是,这是行不通的.

获取已检查的节点.

$("#" +div2.childNodes[i].id).jstree("get_checked",null,true).each 
      (function () { 
         alert(this.data);
         alert(this.id);

 }); 

请给我一个解决方案.

Answer 1

正如jstree的作者(Ivan Bozhanov)指出关于get_checked的google-Groups讨论,它也可以使用以下方法实现:

$('#tree').jstree(true).get_selected();

这将返回ID的列表,例如["j1_2"]或["j1_2","j1_3","j1_1"]

查看Ivan Bozhanov自己的小提琴:jsfiddle-example get_selected

Answer 2

  function submitMe(){ 
        var checked_ids = []; 
        $("#server_tree").jstree("get_checked",null,true).each 
            (function () { 
                checked_ids.push(this.id); 
            }); 
           doStuff(checked_ids); 

通过这个曾经 jstree谷歌组

Answer 3

$.each($("#jstree_demo_div").jstree("get_checked",true),function(){alert(this.id);});

Answer 4

$('#dvTreeStructure').on('changed.jstree', function (e, data) {
                var i, j, r = [];
                for (i = 0, j = data.selected.length; i < j; i++) {
                    r.push(data.instance.get_node(data.selected[i]).text.trim());
                }
                alert('Selected: ' + r.join(', '));

            }