节点树展开后如何点击根节点不展开右键菜单
设置了禁止选中父节点<div class="mini-fit">
<ul id="unitTree" class="mini-tree" url="" style="height:100%;padding:5px;"
contextMenu="#treeMenu" showTreeIcon="true" textField="text" idField="id" parentField="pid" resultAsTree="false" expandOnLoad="false" >
</ul>
<ul id="treeMenu" class="mini-contextmenu" >
<li name="add" id="add" iconcls="icon-add">新增</li>
<li name="remove" id="remove" iconcls="icon-remove">删除</li>
<li name="edit" id="editRole" iconcls="icon-edit">修改</li>
</ul>
</div>
function beforenodeselect(e) { //禁用此方法
//禁止选中父节点
if (e.isLeaf == false) e.cancel = true;
}
function onBeforeOpen(e) {
var menu = e.sender;
var node = tree.getSelectedNode();
if (!node) {
e.htmlEvent.preventDefault();//阻止浏览器默认右键菜单
e.cancel = true;
return;
}
}
在选中子节点后,右键单击父节点还是会显示右键菜单,怎么解决
onbeforeopen里面判断是不是父节点,如果是父节点就e.cancel=true阻止掉
var isLeaf=tree.isLeaf(node) dforce 发表于 2018-4-11 13:56 static/image/common/back.gif
onbeforeopen里面判断是不是父节点,如果是父节点就e.cancel=true阻止掉
var isLeaf=tree.isLeaf(node) ...
谢谢:P,问题解决了
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