c# - 删除节点的 Xpath 查询

Question

我有这样的xml文件的soap序列化部分，现在我想要删除一些对象ex：我想要带有名称组件的图像

<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>

所以请给我一些想法如何实现这一点，我已经尝试了几种我未能实现的方法，请帮助我......提前致谢

score 0 · Accepted Answer

您可以像这样使用 XSLT：

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

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

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（XDocument）：

var doc = XDocument.Parse(xml);

var navigator = doc.CreateNavigator();
var manager = new XmlNamespaceManager(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:Image", manager);

foreach (XPathNavigator node in sel)
{
    Console.WriteLine(node.OuterXml);
}

XPath (XSLT)：

<xsl:template match="*">
    <xsl:for-each select="/descendant::spo:Image">
        <xsl:copy>
            <xsl:copy-of select="Name"/>
        </xsl:copy>
    </xsl:for-each>
</xsl:template>

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);
    }
}

score 0 · Accepted Answer

这对我有用：

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

将此库用于 XPath：https ://github.com/ChuckSavage/XmlLib/

score 0 · Accepted Answer

使用这个 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>

c# - 删除节点的 Xpath 查询

3 回答 3

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