单击父节点时检查树的子节点 [ExtJS] [英] check child nodes of a tree when a parent is clicked [ExtJS]

问题描述

我想知道如何在单击 ExtJs 中的特定节点时检查树的兄弟节点.

I would like to know how can i check the sibling nodes of a tree while clicking on a particular node in ExtJs.

我已经为每个节点提供了 id，我可以访问被点击节点的 id.那么我怎样才能继续自动检查子节点??

I had given id's for each node and i can access the id of a clicked node. then how can i proceed to checking the child nodes automatically ??

有人请帮我..

推荐答案

// or any other way of getting hands on the node you want to work with
var node = treePanel.getNodeById('your-id');
node.eachChild(function(n) {
    n.getUI().toggleCheck(true);
});

如果你想让它在当前节点的整个子树上工作，你必须做一些递归.

If you want this to work on the whole subtree of the current node, you'll have to do some recursion.

更加集成:

treePanel.on('checkchange', function(node, checked) {
    node.eachChild(function(n) {
        n.getUI().toggleCheck(checked);
    });
});

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