想從字符串參數

Question

獲取XML值我有以下字符串的XML可變

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<person> 
    <first-name>RaJeEv(๏๏)</first-name> 
    <last-name>Diboliya</last-name> 
    <headline>Software Engineer at FASTTRACK INDIA.</headline> 
    <site-standard-profile-request> 
    <url>http://www.linkedin.com/profile?viewProfile=&amp;url> 
    </site-standard-profile-request> 
</person> 

現在我想從這個字符串得到的姓氏和名字。我怎樣才能做到這一點？

Answer 1

例如

public class Program { 
    public static void Main(String[] args) { 
     XDocument xdoc = XDocument.Parse(@"<?xml version=""1.0"" encoding=""UTF-8""  standalone=""yes""?> 
<person> 
    <first-name>RaJeEv(๏๏)</first-name> 
    <last-name>Diboliya</last-name> 
    <headline>Software Engineer at FASTTRACK INDIA.</headline> 
    <site-standard-profile-request> 
    <url>http://www.linkedin.com/profile?viewProfile</url> 
    </site-standard-profile-request> 
</person>"); 

     XElement xe = xdoc.Elements("person").First(); 

     Console.WriteLine("{0} {1}", xe.Element("first-name").Value, xe.Element("last-name").Value); 
    }   
} 

Answer 2

權在MSDN

Parse XML with XmlReader

，但如果你有這樣的結構的一類強類型，你還可以看到這個答案如何將其轉換爲XML之間： Send XML String as Response

Answer 3

這是我怎麼會反序列化這一點 -

創建一個具體的領域類Person

[Serializable()] 
public class Person 
{ 
    [System.Xml.Serialization.XmlElementAttribute("first-name")] 
    public string FirstName{ get; set; } 

    [System.Xml.Serialization.XmlElementAttribute("last-name")] 
    public string LastName{ get; set; } 

    [System.Xml.Serialization.XmlElementAttribute("headline")] 
    public string Headline{ get; set; } 

    [System.Xml.Serialization.XmlElementAttribute("site-standard-profile-request")] 
    public string ProfileRequest{ get; set; } 
} 

使用的XmlSerializer將其轉換爲Person類型

XmlSerializer serializer = new XmlSerializer(typeof(Person)); 
var person = serializer.Deserialize(xml) as Person; 

的屬性就可以像

var firstName = person.FirstName; 
var lastName = person.LastName; 

Answer 4

var person = XElement.Parse(yourString).Element("person"); 
string firstName = person.Element("first-name").Value; 
string lastName = person.Element("last-name").Value; 

Answer 5

訪問這就是你尋找..

 XmlDocument xmldoc = new XmlDocument(); 
     XmlNodeList xmlnode; 
     FileStream fs = new FileStream(xmlFilePath, FileMode.Open, FileAccess.Read); 
     xmldoc.Load(fs); 

     xmlnode = xmldoc.GetElementsByTagName("first-name"); 
     string firstname= string.Empty; 
     if(xmlnode!=null) 
      strOption = Regex.Replace(xmlnode[0].InnerText, @"\t|\n|\r| ", ""); 

     xmlnode = xmldoc.GetElementsByTagName("last-name"); 
     string lastname= string.Empty; 
     if(xmlnode!=null) 
      strOption = Regex.Replace(xmlnode[0].InnerText, @"\t|\n|\r| ", ""); 

希望它能幫助:)