blocks|key|1408530|text|子代将执行您所需的操作，但请确保已将名称空间名称与元素名称一起包含在内。如果你省略它，你可能会得到一个空的列表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1408531|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Descendants will do exactly what you need, but be sure that you have included a namespace name together with element's name. If you omit it, you will probably get an empty list.

blocks|key|1623273|text|下面是一个表示命名空间的示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1623274|String+TheDocumentContent+=
@"
<TheNamespace:root+xmlns:TheNamespace+=+'http://www.w3.org/2001/XMLSchema'+>
+++<TheNamespace:GrandParent>
++++++<TheNamespace:Parent>
+++++++++<TheNamespace:Child+theName+=+'Fred'++/>
+++++++++<TheNamespace:Child+theName+=+'Gabi'++/>
+++++++++<TheNamespace:Child+theName+=+'George'/>
+++++++++<TheNamespace:Child+theName+=+'Grace'+/>
+++++++++<TheNamespace:Child+theName+=+'Sam'+++/>
++++++</TheNamespace:Parent>
+++</TheNamespace:GrandParent>
</TheNamespace:root>
";

XDocument+TheDocument+=+XDocument.Parse(+TheDocumentContent+);

//Example+1:
var+TheElements1+=
from
++++AnyElement
in
++++TheDocument.Descendants(+"{http://www.w3.org/2001/XMLSchema}Child"+)
select
++++AnyElement;

ResultsTxt.AppendText(+TheElements1.Count().ToString()+);

//Example+2:
var+TheElements2+=
from
++++AnyElement
in
++++TheDocument.Descendants(+"{http://www.w3.org/2001/XMLSchema}Child"+)
where
++++AnyElement.Attribute(+"theName"+).Value.StartsWith(+"G"+)
select
++++AnyElement;

foreach+(+XElement+CurrentElement+in+TheElements2+)
{
++++ResultsTxt.AppendText(+"\r\n"+%2B+CurrentElement.Attribute(+"theName"+).Value+);
}|code-block|syntax|javascript|1623275|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|$]]

An example indicating the namespace:

<pre><code>String TheDocumentContent =
@"
&lt;TheNamespace:root xmlns:TheNamespace = 'http://www.w3.org/2001/XMLSchema' &gt;
 &lt;TheNamespace:GrandParent&gt;
 &lt;TheNamespace:Parent&gt;
 &lt;TheNamespace:Child theName = 'Fred' /&gt;
 &lt;TheNamespace:Child theName = 'Gabi' /&gt;
 &lt;TheNamespace:Child theName = 'George'/&gt;
 &lt;TheNamespace:Child theName = 'Grace' /&gt;
 &lt;TheNamespace:Child theName = 'Sam' /&gt;
 &lt;/TheNamespace:Parent&gt;
 &lt;/TheNamespace:GrandParent&gt;
&lt;/TheNamespace:root&gt;
";

XDocument TheDocument = XDocument.Parse( TheDocumentContent );

//Example 1:
var TheElements1 =
from
 AnyElement
in
 TheDocument.Descendants( "{http://www.w3.org/2001/XMLSchema}Child" )
select
 AnyElement;

ResultsTxt.AppendText( TheElements1.Count().ToString() );

//Example 2:
var TheElements2 =
from
 AnyElement
in
 TheDocument.Descendants( "{http://www.w3.org/2001/XMLSchema}Child" )
where
 AnyElement.Attribute( "theName" ).Value.StartsWith( "G" )
select
 AnyElement;

foreach ( XElement CurrentElement in TheElements2 )
{
 ResultsTxt.AppendText( "\r\n" + CurrentElement.Attribute( "theName" ).Value );
}
</code></pre>

blocks|key|1623325|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1623326|xml.Descendants().Where(p+=>+p.Name.LocalName+==+"Name+of+the+node+to+find")|code-block|syntax|javascript|1623327|其中xml是一个XDocument。|offset|length|style|CODE|1623328|请注意，属性Name返回一个具有LocalName和Namespace的对象。这就是为什么如果要按名称进行比较，就必须使用Name.LocalName。|1623329|entityMap^0|0|0|2|3|8|9|0|6|4|G|9|Q|9|1P|E|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]|$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|P|$]]

You can do it this way:

<pre><code>xml.Descendants().Where(p =&gt; p.Name.LocalName == "Name of the node to find")
</code></pre>

where <code>xml</code> is a <code>XDocument</code>.

Be aware that the property <code>Name</code> returns an object that has a <code>LocalName</code> and a <code>Namespace</code>. That's why you have to use <code>Name.LocalName</code> if you want to compare by name.

blocks|key|1408678|text|我使用的是与XmlDocument.SelectNodes方法相同的XPathSelectElements扩展方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1408679|using+System;
using+System.Xml.Linq;
using+System.Xml.XPath;+//+for+XPathSelectElements

namespace+testconsoleApp
{
++++class+Program
++++{
++++++++static+void+Main(string[]+args)
++++++++{
++++++++++++XDocument+xdoc+=+XDocument.Parse(
++++++++++++++++@"<root>
++++++++++++++++++++<child>
++++++++++++++++++++++++<name>john</name>
++++++++++++++++++++</child>
++++++++++++++++++++<child>
++++++++++++++++++++++++<name>fred</name>
++++++++++++++++++++</child>
++++++++++++++++++++<child>
++++++++++++++++++++++++<name>mark</name>
++++++++++++++++++++</child>
+++++++++++++++++</root>");

++++++++++++foreach+(var+childElem+in+xdoc.XPathSelectElements("//child"))
++++++++++++{
++++++++++++++++string+childName+=+childElem.Element("name").Value;
++++++++++++++++Console.WriteLine(childName);
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|1408680|entityMap^0|6|N|Y|J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

I am using <code>XPathSelectElements</code> extension method which works in the same way to <code>XmlDocument.SelectNodes</code> method:

<pre><code>using System;
using System.Xml.Linq;
using System.Xml.XPath; // for XPathSelectElements

namespace testconsoleApp
{
 class Program
 {
 static void Main(string[] args)
 {
 XDocument xdoc = XDocument.Parse(
 @"&lt;root&gt;
 &lt;child&gt;
 &lt;name&gt;john&lt;/name&gt;
 &lt;/child&gt;
 &lt;child&gt;
 &lt;name&gt;fred&lt;/name&gt;
 &lt;/child&gt;
 &lt;child&gt;
 &lt;name&gt;mark&lt;/name&gt;
 &lt;/child&gt;
 &lt;/root&gt;");

 foreach (var childElem in xdoc.XPathSelectElements("//child"))
 {
 string childName = childElem.Element("name").Value;
 Console.WriteLine(childName);
 }
 }
 }
}
</code></pre>

blocks|key|1408729|text|根据@Francisco+Goldenstein的回答，我编写了一个扩展方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1408730|using+System.Collections.Generic;
using+System.Linq;
using+System.Xml.Linq;

namespace+Mediatel.Framework
{
++++public+static+class+XDocumentHelper
++++{
++++++++public+static+IEnumerable<XElement>+DescendantElements(this+XDocument+xDocument,+string+nodeName)
++++++++{
++++++++++++return+xDocument.Descendants().Where(p+=>+p.Name.LocalName+==+nodeName);
++++++++}
++++}
}|code-block|syntax|javascript|1408731|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|$]]

Following @Francisco Goldenstein answer, I wrote an extension method

<pre><code>using System.Collections.Generic;
using System.Linq;
using System.Xml.Linq;

namespace Mediatel.Framework
{
 public static class XDocumentHelper
 {
 public static IEnumerable&lt;XElement&gt; DescendantElements(this XDocument xDocument, string nodeName)
 {
 return xDocument.Descendants().Where(p =&gt; p.Name.LocalName == nodeName);
 }
 }
}
</code></pre>

I have an <code>XDocument</code> object. I want to query for elements with a particular name at any depth using LINQ.
When I use <code>Descendants(&quot;element_name&quot;)</code>, I only get elements that are direct children of the current level. I'm looking for the equivalent of &quot;//element_name&quot; in XPath...should I just use <code>XPath</code>, or is there a way to do it using LINQ methods?

Query an XDocument for elements by name at any depth

我有一个XDocument对象。我想使用LINQ在任何深度查询具有特定名称的元素。当我使用Descendants("element_name")时，我只获取当前级别的直接子元素。我正在寻找"//element_name“在XPath...should中的等价物我只使用XPath，或者有没有办法使用LINQ方法来实现？

问按任意深度按名称查询元素的XDocument
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