使用C＃2.0讀取XML字符串

Question

我有一個XML字符串，使用C＃2.0，我必須讀取該字符串並形成一個鍵值對或單獨的列表。我必須使用下面的XML來進行Web Service的字段映射。

下面

是我的我的XML的樣品

<?xml version="1.0" encoding="utf-8" ?> 
<Integration> 
    <FiledMappings name ="Employee"> 
    <Field Name="EmployeeID"> 
     <DataSource>EmployeeNO</DataSource> 
    </Field> 
    <Field Name="Department"> 
     <DataSource>Department</DataSource> 
    </Field> 
    <Field Name="EmployeeName"> 
     <DataSource>Name</DataSource> 
    </Field> 
    </FiledMappings> 
</Integration> 

Answer 1

試試這個代碼;我用Dictionary和XmlDocument：

var keyValues = new Dictionary<string, string>(); 

var document = new XmlDocument(); 
document.LoadXml(stringXml); 
foreach (XmlNode node in document.SelectNodes(@"//Field")) 
{ 
    keyValues.Add(node.Attributes["Name"].InnerText, 
        node.InnerText); 
} 

Answer 2

你可以使用一個disctionary例如下面

private static IDictionary<string, string> parseReplicateBlock(StreamReader reader) 

Answer 3

你可以使用下面的代碼來獲得所需要的詞典：

 StringBuilder s = new StringBuilder(); 
     s.AppendLine("<?xml version=\"1.0\" encoding=\"utf-8\" ?>"); 
     s.AppendLine("<Integration>"); 
     s.AppendLine("<FiledMappings name =\"Employee\">"); 
     s.AppendLine("<Field Name=\"EmployeeID\">"); 
     s.AppendLine("<DataSource>EmployeeNO</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("<Field Name=\"Department\">"); 
     s.AppendLine("<DataSource>Department</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("<Field Name=\"EmployeeName\">"); 
     s.AppendLine("<DataSource>Name</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("</FiledMappings>"); 
     s.AppendLine("</Integration>"); 

     Dictionary<string, string> d = new Dictionary<string, string>(); 

     XmlDocument doc = new XmlDocument(); 
     doc.LoadXml(s.ToString()); 

     XmlNode x = doc.ChildNodes[1].ChildNodes[0]; 
     foreach (XmlNode n in x.ChildNodes) 
      d[n.Attributes[0].Value] = n.FirstChild.FirstChild.Value; 

     foreach (KeyValuePair<string, string> p in d) 
      Console.WriteLine(string.Format("{0}:\t{1}", p.Key, p.Value)); 

     Console.ReadLine(); 

或者，如果它可能似乎可以使用.net 3.5，您可以利用Linq到XML，請參閱：

 StringBuilder s = new StringBuilder(); 
     s.AppendLine("<?xml version=\"1.0\" encoding=\"utf-8\" ?>"); 
     s.AppendLine("<Integration>"); 
     s.AppendLine("<FiledMappings name =\"Employee\">"); 
     s.AppendLine("<Field Name=\"EmployeeID\">"); 
     s.AppendLine("<DataSource>EmployeeNO</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("<Field Name=\"Department\">"); 
     s.AppendLine("<DataSource>Department</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("<Field Name=\"EmployeeName\">"); 
     s.AppendLine("<DataSource>Name</DataSource>"); 
     s.AppendLine("</Field>"); 
     s.AppendLine("</FiledMappings>"); 
     s.AppendLine("</Integration>"); 

     XElement x = XElement.Parse(s.ToString()); 

     Dictionary<string, string> d = x.Element("FiledMappings").Elements("Field").ToDictionary(e => e.Attribute("Name").Value, e => e.Element("DataSource").Value); 
     foreach (KeyValuePair<string, string> p in d) 
      Console.WriteLine(string.Format("{0}:\t{1}", p.Key, p.Value)); 

     Console.ReadLine(); 

Answer 4

這是我能夠做動態：

private Dictionary<string, string> Load_XML_wsMapping(String _WSMapping) 
     { 
      // Web Service Mapping forms Key Value Pair 
      XmlDocument doc = new XmlDocument(_WSMapping); 
      doc.LoadXml(); 
      Dictionary<string, string> dictXMLMapping = new Dictionary<string, string>(); 

      try 
      { 
       XmlNodeList list = doc.FirstChild.NextSibling.FirstChild.ChildNodes; 

       for (int i = 0; i < list.Count; i++) 
       { 
        XmlElement el = ((System.Xml.XmlElement)(list[i])); 
        dictXMLMapping.Add(el.Attributes[0].Value, list[i].InnerText); 

       } 
      } 
      catch (Exception err) 
      { 

       throw new Exception("Error occurred while mapping Web Service -- Bad XML." + err.Message); 
      } 
      return dictXMLMapping ; 
     }