javascript - 我怎么知道所选节点是否是树中的叶子？ (ExtJS 4)

Question

我有一个树面板，我想将一个 Action 与点击这棵树的叶子相关联。我设法将一个与树元素上的每次点击相关联，但我不仅与树叶相关联。

这是我的代码，我尝试了非常简单的功能:(菜单是扩展 tree.Panel 的 View )

Ext.define('WSR.controller.MenuControl', {

    extend: 'Ext.app.Controller',

    views: [
        'Menu',
        'Browser'
    ],

    stores: [ 'UserStore' ],

    init: function() {

        this.control({
            'menu': {
                itemclick: this.onItemClicked
            }
        });
    },

    onItemClicked: function() {
        console.log('ItemClicked');
    }
});

有人帮我吗？

Answer 1

尝试:

Ext.define('WSR.controller.MenuControl', {

extend: 'Ext.app.Controller',

views: [
    'Menu',
    'Browser'
],

refs:[
    {
        selector: 'Menu', //Suggest u to use a selector that is more "tightened" in case you have more trees of this type
        ref: 'mynicetree'
    }
],

stores: [ 'UserStore' ],

init: function() {

    this.control({
        'menu': {
            //I suggest u use selectionchange instead, itemclick won't work if the use navigate the tree nodes using keyboard, whereas selectionchange fit both
            selectionchange: this.loadStuff
        }
    });
},

loadStuff: function() {

    //Assuming you can only select one node at a time (typical use case)
    var myTreePanel = this.getMynicetree();
    var nodes = myTreePanel.getSelectionModel().selection();
    if(!Ext.isEmpty(nodes))
    {
        var selNode = nodes[0];
        if(selNode.get("leaf")) //Assuming your NodeInterface "leaf" property was assigned
        {
            //Its a leaf
        }

        //OR
        if(selNode.isLeaf())
        {
            //Its a leaf
        }

    }
}

});

喂!