blocks|key|999802|text|我今天就是这样做的，下面的代码应该可以工作(使用umbraco.presentation.nodeFactory)，使用nodeId+-1调用它来获取站点的根节点，并让它工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|999803|private+void+DoSomethingWithAllNodesByType(int+NodeId,+string+typeName)
{
++++var+node+=+new+Node(nodeId);
++++foreach+(Node+childNode+in+node.Children)
++++{
++++++++var+child+=+childNode;
++++++++if+(child.NodeTypeAlias+==+typeName)
++++++++{
++++++++++++//Do+something
++++++++}

++++++++if+(child.Children.Count+>+0)
++++++++++++GetAllNodesByType(child,+typeName);
++++}
}|code-block|syntax|javascript|999804|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I was just doing this today, something like the below code should work (using umbraco.presentation.nodeFactory), call it with a nodeId of -1 to get the root node of the site and let it work it's way down:

<pre><code>private void DoSomethingWithAllNodesByType(int NodeId, string typeName)
{
 var node = new Node(nodeId);
 foreach (Node childNode in node.Children)
 {
 var child = childNode;
 if (child.NodeTypeAlias == typeName)
 {
 //Do something
 }

 if (child.Children.Count &gt; 0)
 GetAllNodesByType(child, typeName);
 }
}
</code></pre>

blocks|key|785636|text|或者是递归方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|785637|using+umbraco.NodeFactory;

private+static+List<Node>+FindChildren(Node+currentNode,+Func<Node,+bool>+predicate)
{
++++List<Node>+result+=+new+List<Node>();

++++var+nodes+=+currentNode
++++++++.Children
++++++++.OfType<Node>()
++++++++.Where(predicate);
++++if+(nodes.Count()+!=+0)
++++++++result.AddRange(nodes);

++++foreach+(var+child+in+currentNode.Children.OfType<Node>())
++++{
++++++++nodes+=+FindChildren(child,+predicate);
++++++++if+(nodes.Count()+!=+0)
++++++++++++result.AddRange(nodes);
++++}
++++return+result;
}

void+Example()
{
++++var+nodes+=+FindChildren(new+Node(-1),+t+=>+t.NodeTypeAlias.Equals("myDocType"));
++++//+Do+something...
}|code-block|syntax|javascript|785638|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Or a recursive approach:

<pre><code>using umbraco.NodeFactory;

private static List&lt;Node&gt; FindChildren(Node currentNode, Func&lt;Node, bool&gt; predicate)
{
 List&lt;Node&gt; result = new List&lt;Node&gt;();

 var nodes = currentNode
 .Children
 .OfType&lt;Node&gt;()
 .Where(predicate);
 if (nodes.Count() != 0)
 result.AddRange(nodes);

 foreach (var child in currentNode.Children.OfType&lt;Node&gt;())
 {
 nodes = FindChildren(child, predicate);
 if (nodes.Count() != 0)
 result.AddRange(nodes);
 }
 return result;
}

void Example()
{
 var nodes = FindChildren(new Node(-1), t =&gt; t.NodeTypeAlias.Equals("myDocType"));
 // Do something...
}
</code></pre>

blocks|key|785706|text|假设您最终只需要几个特定类型的节点，那么使用yield关键字可以更有效地避免检索多于必须检索的内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|785707|public+static+IEnumerable<INode>+GetDescendants(this+INode+node)
{
++++foreach+(INode+child+in+node.ChildrenAsList)
++++{
++++++++yield+return+child;

++++++++foreach+(INode+grandChild+in+child.GetDescendants())
++++++++{
++++++++++++yield+return+grandChild;
++++++++}
++++}
++++yield+break;
}|code-block|syntax|javascript|785708|因此，按类型获取节点的最终调用将是：|785709|new+Node(-1).GetDescendants().Where(x+=>+x.NodeTypeAlias+==+"myNodeType")|785710|因此，如果你只想得到前5个结果，你可以在最后加上.Take(5)，这样你就只会递归前5个结果，而不是把整个树都拉出来。|785711|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Supposing you only eventually need a couple of nodes of the particular type, it would be more efficient to use the yield keyword to avoid retrieving more than you have to:

<pre><code>public static IEnumerable&lt;INode&gt; GetDescendants(this INode node)
{
 foreach (INode child in node.ChildrenAsList)
 {
 yield return child;

 foreach (INode grandChild in child.GetDescendants())
 {
 yield return grandChild;
 }
 }
 yield break;
}
</code></pre>

So your final call to get nodes by type will be:

<pre><code>new Node(-1).GetDescendants().Where(x =&gt; x.NodeTypeAlias == "myNodeType")
</code></pre>

So if you only want to get the first five, you can add .Take(5) to the end and you will only recurse through the first 5 results rather than pull out the whole tree.

blocks|key|999999|text|如果你只是在创建一个宏要使用的剃刀脚本文件(Umbraco+4.7%2B)，我发现这个速记特别有用...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1000000|var+nodes+=+new+Node(-1).Descendants("DocType").Where("Visible");|code-block|syntax|javascript|1000001|希望这对某些人有帮助！|1000002|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you're just creating a razor scripting file to be used by a macro (Umbraco 4.7+), I found this shorthand particularly useful...

<pre><code>var nodes = new Node(-1).Descendants("DocType").Where("Visible");
</code></pre>

Hope this helps somebody!

blocks|key|785927|text|在it+7.0%2B中，你可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|785928|foreach+(var+childNode+in+node.Children<ChildNodeType>())
{
...
}|code-block|syntax|javascript|785929|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In umbraco 7.0+ you can do it like this

<pre><code>foreach (var childNode in node.Children&lt;ChildNodeType&gt;())
{
...
}
</code></pre>

Using Umbraco 4.6+, is there a way to retrieve all nodes of a specific doctype in C#? I've been looking in the <code>umbraco.NodeFactory</code> namespace, but haven't found anything of use yet.

Umbraco 4.6+ - How to get all nodes by doctype in C#?

有没有一种方法可以在C#中检索特定文档类型的所有节点？我一直在寻找umbraco.NodeFactory名称空间，但还没有找到任何有用的东西。

问如何在C#中通过4.6+获取所有节点？
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问如何在C#中通过4.6+获取所有节点？
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