How to read the xml file

sir,
i used the following query, but it shows me the error. plz help wer i made a mistake
public void fixtures()
    {
      XElement fixture = null;
      try
      {
        fixture = XElement.Load("http://synd.cricbuzz.com/onmobile-togo/schedule-internal.xml");
      }
      catch
      {
      }
      if (fixture != null)
      {
        var fix = from futu in fixture.Descendants("togoText")
            select (new{
              tit = futu.Element("title").Value,
              par = futu.Element("p").Value
            });
        Repeater1.DataSource = fix;
        Repeater1.DataBind();
      }
    }



    ,
    ,
   

sir... watch this below link,
http://synd.cricbuzz.com/ashwl/scores-multi.xml and http://synd.cricbuzz.com/score-gadget/gadget-scores-feed.xml
from this link i am getting live score.

I doubt it is RSS, since RSS has specific tags and the tags I see in the data you showed are not RSS tags.

hi sir,again doubt in the same xml. in my xml file the 1st child has some sub child and the 2nd child has some extra sub child elements.
<matches>
<match id="20110825-ireeng">



its a rss xml file, so changes occurs oftenly. some times both child have the same elements.
my need is i want to display it dynamically. please help me...

Define templates in Repeater like this:


The Eval() function will bind the properties you have created in Linq to individual labels.

hi Amit,

i used the following query
var xml2 = from s3 in hell.Descendants("match") select (new{
                Label3 = s3.Element("team1").Value,
                Label4 = s3.Element("team2").Value,
                Label5=s3.Element("series").Value
            });
and bind it to repeater
Repeater2.DataSource = xml2;
            Repeater2.DataBind();
          
my output is
i need without this curly braces and i want to directly the value to label in repeater.please suggest me an idea

Note that there is also something called Extensible Stylesheet Language Transformations (XSLT) that can be used to format XML without writing any code. Well some people would consider XSL to be code. XSLT is complex and therefore very useful and the basics can be learned relatively quickly but it does take time to learn. It is rarely discussed in these forums but it might be better for you than LINQ. You can find very many books and other articles that can help. The advantage of XSL for you is that it can do the selection and formatting; both in one thing.

ok chitra.. first of all to become a good developer you should be very curious to things,always ready to learn, understand the problems, always thinking of better code.. and never trust your code, keep testing it;)

For your learning i suggest you to read the latest articles in this site.. this is your site, search it, explore it, learn it and don't forget to say thanks to authors. 
These days in .Net LINQ is something every developer should good at it.
List the items you want to learn and start reading and do some sample project even small ones.
LINQ (includes Anonymous types, Lambda expressions, Extensions methods, Var keyword, Initializers etc. )
LINQ to Objects
LINQ to XML
LINQ to SQL (this is sort of outdated but learning it will be addon as you will know wat's new in Entities which was not in Linq to Sql)
LINQ to Entities

Prepare your self... 

thank u amit, and i am new to developing, so can u plz tell me...how to learn linq and xml etc.. i am .net developer, and wat r all i should learn to become programmer.

amit, now that link displays 2 msgs, so it shows error, i use the following code

public void getscoress()
    {
        XmlDocument xml = new XmlDocument();
        xml.LoadXml("http://synd.cricbuzz.com/score-gadget/gadget-scores-feed.xml"); // suppose that myXmlString contains "..."

        XmlNodeList xnList = xml.SelectNodes("/matches/match");
        foreach (XmlNode xn in xnList)
        {
            string score = xn["header"].InnerText;
            string des = xn["description"].InnerText;
            Console.WriteLine("score: {0} {1}", score, des);
        }
    },
it shows the error like this Data at the root level is invalid. Line 1, position 1.

Hi Chitra,

If amit's answer solved your query then Mark as Acceped Answer.

Thanks

Hi Chitra,

use below code:



Output:
 live scores: ZIM 191/2 (60.0 ovs)

Hope this was helpful.