How to read the xml file

Question

Hi sir,

I want to display the live cricket score in my web application, the following link gives the score
http://synd.cricbuzz.com/score-gadget/gadget-scores-feed.xml

I want to display this score in labels r any gridview control.
plz help me...

Amit Choudhary · Answer

Hi Chitra, if you want the very first item then change the Linq query to var xml = ( from score in xmlEle.Descendants( "match" ) select score).FirstOrDefault(); or if you want to show all the matches information then take a data control and bind it by selecting all the records. Remove the FirstOrDefault() all elements would get selected. moreover you can specify the new type in select of linq: like this: var xml2 = from score in xmlEle.Descendants( "match" ) select ( new {Header= score.Element( "header" ).Value, Description=score.Element( "description" ).Value, UrlText=score.Element( "url-text" ).Value, UrlLink=score.Element( "url-link" ).Value }); Bind these properties to the data control. Note: Take care of data in each element of original XML file it should be in correct form. Hope this was helpful.

Amit Choudhary · Answer

Hi Chitra, use below code: using System.Xml.Linq; void GetLiveScore() { XElement xmlEle= null ; try { xmlEle = XElement .Load( "http://synd.cricbuzz.com/score-gadget/gadget-scores-feed.xml" ); } catch { } if (xmlEle != null ) { var xml = ( from score in xmlEle.Descendants( "match" ) select score).SingleOrDefault(); if (xml != null ) { lblliveScore.Text = xml.Element( "header" ).Value; } } } } Output: live scores: ZIM 191/2 (60.0 ovs) Hope this was helpful.

chitra · Answer

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(); } } '> , '> ,

chitra · Answer

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.

Sam Hobbs · Answer

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

chitra · Answer

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-susind " > team1 > Sussex team1 > team2 > India team2 > series > India in England 2011 - Practice matches series > startdate > 25 08 2011 startdate > enddate > 25 08 2011 enddate > match id =" 20110825-ireeng " > team1 > Ireland team1 > team2 > England team2 > series > England in Ireland 2011 series > startdate > 25 08 2011 startdate > enddate > 25 08 2011 enddate > starttimeGMT > 09:15 starttimeGMT > prevDayPerGMT > false prevDayPerGMT > actualendGMT > 20110825170644 actualendGMT > type > ODI type > state > complete state > - currentscore > currentinningsno > 2 currentinningsno > inningsno > 2 inningsno > batteamname > Ireland batteamname > bowlteamname > England bowlteamname > batteamruns > 117 batteamruns > batteamwkts > 8 batteamwkts > batteamovers > 23 batteamovers > currentscore > 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...

Amit Choudhary · Answer

Define templates in Repeater like this: asp : Repeater runat ="server"> ItemTemplate > asp : Label ID ="label1" runat ="server" Text =' # Eval("Label3") %> '> asp : Label > , asp : Label ID ="label2" runat ="server" Text =' # Eval("Label4") %> '> asp : Label > , asp : Label ID ="label3" runat ="server" Text =' # Eval("Label5") %> '> asp : Label > ItemTemplate > asp : Repeater > The Eval() function will bind the properties you have created in Linq to individual labels.

chitra · Answer

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 { Label3 = Zimbabwe, Label4 = Bangladesh, Label5 = Bangladesh in Zimbabwe 2011 } { Label3 = Canada, Label4 = Afghanistan, Label5 = Afghanistan in Canada 2011 } i need without this curly braces and i want to directly the value to label in repeater.please suggest me an idea

chitra · Answer

thank u amit

Sam Hobbs · Answer

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.

Amit Choudhary · Answer

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...

chitra · Answer

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.

chitra · Answer

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.

Dinesh Beniwal · Answer

Hi Chitra, If amit's answer solved your query then Mark as Acceped Answer. Thanks

chitra · Answer

hi amit, thank u,its really working.

How to read the xml file

Answers (15)