xml düğümü için

Question

Numune XML metni alın:

<query yahoo:count="1" yahoo:created="2016-03-31T06:43:49Z" yahoo:lang="en-US"> 
    <results> 
     <channel> 
      <item> 
       <yweather:condition code="28" date="Thu, 31 Mar 2016 08:00 AM SAST" temp="58" text="Mostly Cloudy"/> 
      </item> 
     </channel> 
    </results> 
</query> 

Kodu:

string weburl = "https://query.yahooapis.com/v1/public/yql?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20in%20%28select%20woeid%20from%20geo.places%281%29%20where%20text%3D%22Cape%20Town%22%29&format=xml&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys"; 

var xml = await new WebClient().DownloadStringTaskAsync(new Uri(weburl)); 

XmlDocument doc = new XmlDocument(); 
doc.LoadXml(xml); 

XmlElement root = doc.DocumentElement; 

XmlNodeList nodes = root.SelectNodes("//query/results/channel/item"); 

foreach (XmlNode node in nodes) 
{ 
    MessageBox.Show(node.InnerXml); 
} 

Sadece ve metin outputed temp almak için mücadele edilmiştir ama yol bulamıyorum nasıl, bu sahip olduğum kadarıyla. XML erişebilir

Answer 1

XmlNode.Attributes özelliğinden niteliklerini:

var condition = doc.SelectSingleNode("/query/results/channel/item/*"); 
MessageBox.Show(condition.Attributes["text"].Value); 
MessageBox.Show(condition.Attributes["temp"].Value); 

Answer 2

Ben Regex ile birlikte xml linq kullandı. Xml ile ilgili sorunları düzeltmem gerekiyordu. Bence ana konu isim alanlarıydı.

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using System.Xml; 
using System.Xml.Linq; 
using System.Text.RegularExpressions; 

namespace ConsoleApplication1 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      string xml = 
       "<?xml version=\"1.0\" encoding=\"utf-8\" ?>" + 
       "<Root xmlns:yahoo=\"abc\" xmlns:yweather=\"def\">" + 
       "<query yahoo:count=\"1\" yahoo:created=\"2016-03-31T06:43:49Z\">" + 
        "yahoo:lang=\"en-US\"><results>" + 
        "<channel>" + 
         "<item>" + 
         "<yweather:condition>" + 
          "code=\"28\" date=\"Thu, 31 Mar 2016 08:00 AM SAST\" temp=\"58\" text=\"Mostly Cloudy\"/>" + 
         "</yweather:condition>" + 
         "</item>" + 
        "</channel>" + 
        "</results>" + 
       "</query>" + 
       "</Root>"; 

      XDocument doc = XDocument.Parse(xml); 

      string innertext = doc.Descendants().Where(x => x.Name.LocalName == "condition").Select(y => y.Value).FirstOrDefault(); 
      string pattern = "\\s?(?'name'[^=]+)=\"(?'value'[^\"]+)\""; 
      MatchCollection matches = Regex.Matches(innertext, pattern); 
      foreach (Match match in matches) 
      { 
       Console.WriteLine("Name : {0}, Value : {1}", match.Groups["name"].Value, match.Groups["value"].Value); 
      } 
      Console.ReadLine(); 
     } 
    } 
} 

Answer 3

bu deneyin ....

usings ....

using System.IO; 
using System.Net; 
using System.Text; 
using System.Xml; 
using System.Xml.Serialization; 

Sınıflar ....

[XmlRoot(ElementName = "condition", Namespace = "http://xml.weather.yahoo.com/ns/rss/1.0")] 
    public class Condition 
    { 
     [XmlAttribute(AttributeName = "yweather", Namespace = "http://www.w3.org/2000/xmlns/")] 
     public string Yweather { get; set; } 
     [XmlAttribute(AttributeName = "code")] 
     public string Code { get; set; } 
     [XmlAttribute(AttributeName = "date")] 
     public string Date { get; set; } 
     [XmlAttribute(AttributeName = "temp")] 
     public string Temp { get; set; } 
     [XmlAttribute(AttributeName = "text")] 
     public string Text { get; set; } 
    } 

    [XmlRoot(ElementName = "item")] 
    public class Item 
    { 
     [XmlElement(ElementName = "condition", Namespace = "http://xml.weather.yahoo.com/ns/rss/1.0")] 
     public Condition Condition { get; set; } 
    } 

    [XmlRoot(ElementName = "channel")] 
    public class Channel 
    { 
     [XmlElement(ElementName = "item")] 
     public Item Item { get; set; } 
    } 

    [XmlRoot(ElementName = "results")] 
    public class Results 
    { 
     [XmlElement(ElementName = "channel")] 
     public Channel Channel { get; set; } 
    } 

    [XmlRoot(ElementName = "query")] 
    public class Query 
    { 
     [XmlElement(ElementName = "results")] 
     public Results Results { get; set; } 
     [XmlAttribute(AttributeName = "yahoo", Namespace = "http://www.w3.org/2000/xmlns/")] 
     public string Yahoo { get; set; } 
     [XmlAttribute(AttributeName = "count", Namespace = "http://www.yahooapis.com/v1/base.rng")] 
     public string Count { get; set; } 
     [XmlAttribute(AttributeName = "created", Namespace = "http://www.yahooapis.com/v1/base.rng")] 
     public string Created { get; set; } 
     [XmlAttribute(AttributeName = "lang", Namespace = "http://www.yahooapis.com/v1/base.rng")] 
     public string Lang { get; set; } 
    } 

Kod ...

 XmlDocument xmlDocument = new XmlDocument(); 
     try 
     { 
      xmlDocument.Load("https://query.yahooapis.com/v1/public/yql?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20in%20%28select%20woeid%20from%20geo.places%281%29%20where%20text%3D%22Cape%20Town%22%29&format=xml&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys"); 

      string XMLxmlDocument = xmlDocument.InnerXml.ToString(); 
      byte[] BUFXML = ASCIIEncoding.UTF8.GetBytes(XMLxmlDocument); 
      MemoryStream ms1 = new MemoryStream(BUFXML); 

      XmlSerializer DeserializerPlaces = new XmlSerializer(typeof(Query));//, new XmlRootAttribute("Query")); 
      using (XmlReader reader = new XmlTextReader(ms1)) 
      { 
       Query dezerializedXML = (Query)DeserializerPlaces.Deserialize(reader); 

       string temp = dezerializedXML.Results.Channel.Item.Condition.Temp; 
       string text = dezerializedXML.Results.Channel.Item.Condition.Text; 

      }// Put a break-point here, then mouse-over temp and text, you should have you values (dezerializedXML contains the entire object) 
     } 
     catch (System.Exception) 
     { 
      throw; 
     }