用于删除节点的Xpath查询

Question

我有像这样的xml文件的soap序列化的一部分，现在我想要删除一些对象例如:我想图像与名称组件

<SOAP-ENV:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:clr="http://schemas.microsoft.com/soap/encoding/clr/1.0" SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/">
<SOAP-ENV:Body>
<a1:Image id="ref-1" xmlns:a1="http://schemas.microsoft.com/clr/nsassem/Spo.DataModel/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58">
<Name id="ref-4">Component</Name>
<ImmediateState xsi:type="a2:ModifiedState" xmlns:a2="http://schemas.microsoft.com/clr/nsassem/Spo.Plugins/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58">Current</ImmediateState>
</a1:Image>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>
<SOAP-ENV:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:clr="http://schemas.microsoft.com/soap/encoding/clr/1.0" SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/">
<SOAP-ENV:Body>
<a1:Image id="ref-1" xmlns:a1="http://schemas.microsoft.com/clr/nsassem/Spo.DataModel/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58">
<Name id="ref-4">Connect</Name>
<ImmediateState xsi:type="a2:ModifiedState" xmlns:a2="http://schemas.microsoft.com/clr/nsassem/Spo.Plugins/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58">Current</ImmediateState>
</a1:Image>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>

所以请给我提供一些如何实现这一点的想法，我已经尝试了几种方法，但都失败了，请帮助我…提前感谢

Answer 1

1. 您可以这样使用XSLT：

>

1. Linq2Xml：

var doc = XDocument.Parse(xml)；var elementName = XName.Get( "Image"，"http://schemas.microsoft.com/clr/nsassem/Spo.DataModel/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58")；var items = from x in doc.Descendants().Elements(elementName) select new { ImageId = x.Attribute("id").Value，NameId = x.Element(" Name ").Attribute("id").Value，Name= x.Element("Name").Value };

• XPath ()：

var doc = XDocument.Parse(xml)；var navigator = doc.CreateNavigator()；var manager =新图像(navigator.NameTable)；manager.AddNamespace("spo"，"http://schemas.microsoft.com/clr/nsassem/Spo.DataModel/Spo.DataModel%2C%20Version%3D12.1.3.0%2C%20Culture%3Dneutral%2C%20PublicKeyToken%3D23bd062a94e26d58")；var sel =navigator.Select(“//descendant：：spo:XmlNamespaceManager”，manager)；foreach (sel中的XPathNavigator节点){ Console.WriteLine(node.OuterXml)；}

• XPath (XSLT):

更新

var xml = string.Format(@"<root>{0}</root>", Resource1.String1);
var doc = XDocument.Parse(xml);
// process document
using (var file = File.CreateText(@"file.xml"))
{
    foreach (XElement nodes in doc.Root.Elements())
    {
        file.Write(nodes);
    }
}

Answer 2

这对我很有效：

XElement root = XElement.Load(file);
IEnumerable<XElement> images = root.XPath("//Image[Name]");

将这个库用于XPath：https://github.com/ChuckSavage/XmlLib/

Answer 3

使用此XSLT转换

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="node()|@*">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match="*[local-name() = 'Image' and Name]"/>
</xsl:stylesheet>