javascript 如何将节点标识为根节点？
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How do I identify a node as being a root node?
提问by Eddy Pronk
I want to create a node inside or under an existing node depending on if it is a root node. (a tree widget is usually a list of trees or a tree without a visible root node.)
我想在现有节点内部或下方创建一个节点，具体取决于它是否是根节点。（树小部件通常是树的列表或没有可见根节点的树。）
I tried get_parent, but how do I know if that is a root node?
我试过 get_parent，但我怎么知道那是不是根节点？
var parent = $("#demo1").jstree('_get_parent', $("#foo"));
var node = $("#demo1").jstree('_get_node', $("#foo"));
What confused me is that get_node seems to return the same object as get_parent.
让我感到困惑的是 get_node 似乎返回与 get_parent 相同的对象。
I'm using jstree_pre1.0_fix_1.
我正在使用 jstree_pre1.0_fix_1。
edited:
编辑：
I ended up checking for the known id of the parent of the parent.
我最终检查了父级的父级的已知 ID。
var node = $(e.replyto);
if (node.length) {
if (node.parent().parent().attr('id') == 'demo1') {
$("#demo1").jstree("create_node", node, 'last',{'attr': {'id':e.id}, 'state':'open', 'data': e.data}) ;
} else {
$("#demo1").jstree("create_node", node, 'after',{'attr': {'id':e.id}, 'state':'open', 'data': e.data}) ;
}
} else {
$("#demo1").jstree("create_node", -1, 'after',{'attr': {'id':e.id}, 'state':'open', 'data': e.data});
}
回答by Nada_Surf
You can call get_parent()on the node. If it returns '#' then the node is a root node. E.G:
您可以在节点上调用get_parent()。如果它返回“#”，则该节点是根节点。例如：
var node = ...;
if($('#demo1').jstree(true).get_parent(node) == '#') {
// node is a root node
}
回答by Thomas Guillory
It's not the ideal solution but you can use _get_childrenwith -1in param to get all root nodes and test if your node is in the list.
这不是理想的解决方案，但您可以使用_get_childrenwith -1in param 来获取所有根节点并测试您的节点是否在列表中。
._get_children ( node )
Use -1 to return all root nodes.
回答by DavidHyogo
I've been struggling a bit with this because I'm trying to use the context menu plugin. I don't want the user to be able to create a new root node. I only want to allow them to create sub-nodes. (The root nodes represent the groups that the current user belongs to so is pre-set by the administrator). My first confusion was that _get_childrenreturns an object with a lengthproperty. It's not an array but it has a lengthproperty corresponding correctly to the number of actual root nodes. Looking at the underlying code, the jstree_get_childrenmethod uses jQuery's childrenmethod, which returns the child nodes with indexes 0, 1, 2 etc along with other jQuery properties and methods. I found it convenient to extract just the array of nodes and use indexOfto check if the current node is a root node. So, here is a snippet from the itemsproperty of my jstreecontext menu configuration:
我一直在为此苦苦挣扎，因为我正在尝试使用上下文菜单插件。我不希望用户能够创建新的根节点。我只想让他们创建子节点。（根节点代表当前用户所属的组，由管理员预先设置）。我的第一个困惑是_get_children返回一个带有length属性的对象。它不是一个数组，但它有一个length与实际根节点数正确对应的属性。查看底层代码，该jstree_get_children方法使用 jQuery 的children方法，该方法返回索引为 0、1、2 等的子节点以及其他 jQuery 属性和方法。我发现仅提取节点数组并使用很方便indexOf检查当前节点是否为根节点。所以，这是items我的jstree上下文菜单配置的属性的一个片段：
'items': function(node){
var rootChildren = this._get_children(-1),
rootNodes = [],
i;
//rootChildren is now a fancy jQuery object with the child nodes assigned
//to keys 0, 1 etc.
//Now create a simple array
for(i = 0; i < rootChildren.length; i += 1){
rootNodes[i] = rootChildren[i];
}
//We can now use indexOf to check if the current node is in that array
//Note again that node is a fancy jQuery object, the actual DOM element
//is stored in node[0]
console.log(rootNodes.indexOf(node[0]) > -1);
//code here to add whatever items you want to the context menu.
}
If you right click around your tree, you will see truein the console window for the root nodes and falsefor any nodes lower down the hierarchy. Note that for IE below 8 (I think), you will need to supply an indexOfmethod for Arraybecause earlier versions of IE don't supply indexOfas a native method.
如果您在树上右键单击，您将true在控制台窗口中看到根节点和false层次结构下方的任何节点。请注意，对于低于 8 的 IE（我认为），您需要提供一种indexOf方法，Array因为早期版本的 IE 不indexOf作为本机方法提供。
