Question

我有一个xml文件，我需要解析它以获取信息 -

 <?xml version="1.0" encoding="utf-8" standalone="yes"?>
 <feed xml:base="http://google.com/en-US/syndicate/" xmlns:d="http://schemas.google.com/ado/2007/08/dataservices" xmlns:m="http://schemas.giooglt.com/ado/2007/08/dataservices/metadata" xmlns="http://www.w3.org/2005/Atom">
  <title type="text">Partners</title>
  <id>http://googlre.com/en-US/syndicate/Partners</id>
  <updated>2014-01-16T21:33:20Z</updated>
  < link rel="self" title="Partners" href="Partners" />
<entry>
   <id>http://pinpoint.microsoft.com/en-US/syndicate/Partners('4555')</id>
   <title type="text">M55p; Co</title>
   <summary type="text">cccc is a Certified Partner, reseller, and implementer of 
 Key industries we work with include:
•   Financial services
•   Professional services
•   Media / publishing


 By focusing on mid-market to enterprise clients, 
</summary>
 <published>2009-07-21T14:23:50-07:00</published>
 <updated>2013-11-22T15:00:46-08:00</updated>
 <author>
   <name>google chrome</name>
  <uri>http://google.com/</uri>
  <email>retee@gmail.com</email>
</author>
<link rel="edit" title="Partner" href="Partners('4255')" />
<link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/Links" type="application/atom+xml;type=feed" title="Links" href="Partners('4559')/Links">
  <m:inline>
    <feed>
      <title type="text">Links</title>
      <id>http://google.com/('429')/Links</id>
      <updated>2014-01-16T21:33:20Z</updated>
      <link rel="self" title="Links" href="Partners('4ff')/Links" />
      <entry>
        <id>http://ryryr.com/en-US/syndicate/Links('ufufr')</id>
        <title type="text">
        </title>
        <updated>2014-01-16T21:33:20Z</updated>
        <author>
          <name />
        </author>
        <link rel="edit" title="Link" href="Links('partnerpage')" />
        <category term="google.Commerce.ferrr.Syndicate.V2010_05.Link" scheme="http://schemas.frrr.com/ado/2007/08/dataservices/scheme" />
        <content type="application/xml">
          <m:properties>
            <d:Type>pgooglrpartnerpage</d:Type>
            <d:Description>google Partner Page</d:Description>
            <d:Url>http://googlgt.com/en-US/PartnerDetails.aspx?PartnerId=42555&amp;wt.mc_id=66ttet</d:Url>
          </m:properties>
        </content>
      </entry>
      </m:inline>
      </entry>

我正在使用这段代码生成信息 -

    // Alternate Method for getting the Fields from the XML file
         XmlDocument xmlDocument = new XmlDocument();
         xmlDocument.Load("C:/Users/Administrator/Downloads/direct.xml");
        XmlNamespaceManager xmlnm = new XmlNamespaceManager(xmlDocument.NameTable);
        xmlnm.AddNamespace("de","http://www.w3.org/2005/Atom");

        ParseXML(xmlDocument, xmlnm);

        Debug.WriteLine("\n---XML parsed---");

    }

    public static void ParseXML(XmlDocument xmlFile, XmlNamespaceManager xmlnm)
    {
        //XmlNodeList nodes = xmlFile.SelectNodes("//ns:entry/ns:updated| //ns:entry/ns:published | //ns:entry/ns:id ", xmlnm);
        XmlNodeList nodes = xmlFile.SelectNodes("//de:entry/de:link/de:inline/de:feed/de:id ", xmlnm);

        foreach (XmlNode node in nodes)
        {
            Debug.WriteLine(node.Name + " = " + node.InnerXml);
        }

在上面的程序中，我希望获得的信息为“// de：entry / de：link / de：m：inline / de：feed / de：id”即尝试提取信息，但程序确实如此不承认“/ de：m：inline”作为正确的格式并生成错误，指出“格式不正确”，我希望解析信息以获取指定的数据，但不知道如何生成正确的格式。 C＃的新用户请帮我解决以下问题。

失败的代码示例：

xmlFile.SelectNodes("//de:link/de:m:inline", xmlnm);

错误：

'// de：m：inline'的标记无效

Answer 1

添加名称空间de和m

public static void Main()
    {
        // Alternate Method for getting the Fields from the XML file
        XmlDocument xmlDocument = new XmlDocument();
        xmlDocument.Load("C:/Users/Administrator/Downloads/direct.xml");
        XmlNamespaceManager xmlnm = new XmlNamespaceManager(xmlDocument.NameTable);

        xmlnm.AddNamespace("de", "http://www.w3.org/2005/Atom"); 

        **** add this too****
        xmlnm.AddNamespace("m", "http://schemas.giooglt.com/ado/2007/08/dataservices/metadata");

        ParseXML(xmlDocument, xmlnm);

        Console.WriteLine("\n---XML parsed---");
        Console.ReadKey();

    }



 public static void ParseXML(XmlDocument xmlFile, XmlNamespaceManager xmlnm)
    {
       /// inline should be ""m"" not ""de""
        XmlNodeList nodes = xmlFile.SelectNodes("//de:entry/de:link/m:inline/de:feed/de:id", xmlnm);
        foreach (XmlNode node in nodes)
        {
            Console.WriteLine(node.Name + " = " + node.InnerXml);
        }
    }

Answer 2

要从XML文件中获取信息，您可以使用XDocument将数据传递给它，然后迭代文件中存在的每个元素（节点）并获取其内部节点或属性等。像这样： string PdFileContent = string.Empty;

            string FileContent = string.Empty;

            using (StreamReader Reader = new StreamReader(**FilePath**, UTF8Encoding.UTF8))
            {
                FileContent = Reader.ReadToEnd();
            }

            XmlDocument XDoc = new XmlDocument();

            XDoc.LoadXml(FileContent); //Load the file with XML strcuture

            XDoc.SelectNodes("/**NodeNameToSelect**");

            // Read inner text of the XDocument's Child Elements
            string Result = XDoc["**ChildNodeName**"].InnerText;

如何解析xml文件以获取字段？

2 个答案: