Umbraco - 在 C# 中查找根节点
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Umbraco - Finding Root Node in C#
提问by Atom
I'm working on a backend module, so Node.GetCurrent()is not an option. I need to find a way to call something like Node currentNode = new Node(parentNodeId);and get the root node of the site. I've seen samples in XSLT, but nothing for C#. Does anyone know how I can accomplish this?
我正在研究后端模块，所以Node.GetCurrent()不是一个选项。我需要找到一种方法来调用类似的东西Node currentNode = new Node(parentNodeId);并获取站点的根节点。我在 XSLT 中看到过示例，但没有看到 C# 中的示例。有谁知道我怎么能做到这一点？
Even just getting the ID of the root node so I can call new Node()would be great.
即使只是获取根节点的 ID 以便我可以调用new Node()也会很棒。
采纳答案by E.J. Brennan
The rootnode is always available as:
根节点始终可用：
var rootNode = new Node(-1);
回答by DotNetDan
Brennan is correct,
布伦南是对的，
var rootNode = new DynamicNode(-1);
works as well!
也能用！
回答by tcmorris
Update for Umbraco 6+
Umbraco 6+ 更新
public static IPublishedContent GetRootNode()
{
var umbracoHelper = new UmbracoHelper(UmbracoContext.Current);
var rootNode = umbracoHelper.TypedContentSingleAtXPath("//root"));
return rootNode;
}
This just takes a document type alias and finds the root node as IPublishedContent using the current Umbraco context. UmbracoHelper gives you quite a few options off this also.
这仅需要一个文档类型别名，并使用当前 Umbraco 上下文找到根节点作为 IPublishedContent。UmbracoHelper 也为您提供了很多选择。
回答by tryinHard
Update for Umbraco 7 (may work in earlier versions too)
Umbraco 7 更新（也可以在早期版本中使用）
@{
var siteroot = CurrentPage.AncestorOrSelf(1);
}
For further info, check out the documentation -> http://our.umbraco.org/Documentation/Reference/Querying/DynamicNode/Collections
有关更多信息，请查看文档 -> http://our.umbraco.org/Documentation/Reference/Querying/DynamicNode/Collections
回答by Leszek P
Umbraco 7:
翁布拉科 7：
Umbraco.TypedContentAtRoot();
回答by Gcamara14
I frequently use this one. I like that it's relative so that if you have multiple root nodes you can target both without a foreach loop.
我经常使用这个。我喜欢它是相对的，所以如果你有多个根节点，你可以在没有 foreach 循环的情况下将它们作为目标。
IPublishedContent topNode = Model.Content.AncestorOrSelf(1);
