xml parsing - Read XML Webfile in C# Windows Form and place data in labels -

for information program need read web xml file , place in label every value.

an example of xml file:

<actuelevertrektijden>       <vertrekkendetrein>         <ritnummer>5070</ritnummer>         <vertrektijd>2015-03-20t19:42:00+0100</vertrektijd>         <eindbestemming>den haag centraal</eindbestemming>         <treinsoort>sprinter</treinsoort>         <routetekst>lage zwaluwe, dordrecht, rotterdam c.</routetekst>         <vervoerder>ns</vervoerder>         <vertrekspoor wijziging="false">6</vertrekspoor>      </vertrekkendetrein>      <vertrekkendetrein>         <ritnummer>1971</ritnummer>         <vertrektijd>2015-03-20t19:50:00+0100</vertrektijd>         <eindbestemming>venlo</eindbestemming>         <treinsoort>intercity</treinsoort>         <routetekst>tilburg, eindhoven, helmond</routetekst>         <vervoerder>ns</vervoerder>         <vertrekspoor wijziging="false">4</vertrekspoor>         <opmerkingen> // not available, important , specific 'vertrekkendetrein'           <opmerking>rijdt vandaag niet</opmerking>             </opmerkingen>         </vertrekkendetrein> </actuelevertrektijden> 

i tried xmldocument (using foreach) this:

 foreach (xmlnode nodelist2 in nodelist)             {                 if (i < 1) //1                 {                      switch (nodelist.name)                     {                         case "vertrektijd": string vertrek1 = (nodelist2.innertext); break;                         case "vertrekvertragingstekst": string vertraging1 = (nodelist2.innertext); break;                         case "eindbestemming": string eindbest1 = (nodelist2.innertext); break;                         case "treinsoort": string treinsoort1 = (nodelist2.innertext); break;                         case "routetekst": string route1 = (nodelist2.innertext); break;                         case "vertrekspoor": string spoor1 = (nodelist2.innertext); i++ break;                         case "opmerkingen": case "opmerking": string opmerking1 = (nodelist2.innertext); break;                      }                   }              } 

but wasn't success. there smarter way read it, , place in lot of labels?

edit: answer got, tried following code:

try         {             string urlo = "**secured webaddress not end .xml**";             string resultje = httpgeto(urlo);             xmldocument doc = new xmldocument();              xmlnode root = doc.firstchild;             //* document has 1 root element "actuelevertrektijden"             //* root element has multiple child nodes "vertrekkendetrein"              xmlnodelist nodelist1 = root.childnodes;             (int = 0; < nodelist1.count; i++)             {                 xmlnodelist nodelist2 = nodelist1[i].childnodes;                 //* each child node of of child nodes,                 //* need text within each 1 of them                 foreach (xmlnode node in nodelist2)                 {                     switch (node.name)                     {                         case "vertrektijd":                             string vertrek1 = node.innertext; messagebox.show(vertrek1); lblts1.text = vertrek1;                             break;                         case "vertrekvertragingstekst":                             string vertraging1 = node.innertext;                             break;                         case "eindbestemming":                             string eindbest1 = node.innertext;                             break;                         case "treinsoort":                             string treinsoort1 = node.innertext;                             break;                         case "routetekst":                             string route1 = node.innertext;                             break;                         case "vertrekspoor":                             string spoor1 = node.innertext;                             break;                         case "opmerkingen":                             xmlnode opnode = node.firstchild;                             if (opnode != null)                             {                                 string opmerking1 = opnode.innertext;                             }                             break;                     }                 }             }           }         catch         {             lblcatch.text = "werktniet";                    } 

but doesn't work. , also, how can read next "vertrekkendetrein"?, copying code , use i++;? need read first 6 "vertrekkendetrein".

xmlnode root = doc.firstchild; //* document has 1 root element "actuelevertrektijden" //* root element has multiple child nodes "vertrekkendetrein"  xmlnodelist nodelist1 = root.childnodes; (int i=0; i<nodelist1.count; i++) {   xmlnodelist nodelist2 = nodelist1[i].childnodes;   //* each child node of of child nodes,   //* need text within each 1 of them   foreach (xmlnode node in nodelist2)   {     switch (node.name)     {             case "vertrektijd":    string vertrek1 = node.innertext;    break;   case "vertrekvertragingstekst":    string vertraging1 = node.innertext;    break;   case "eindbestemming":    string eindbest1 = node.innertext;    break;   case "treinsoort":    string treinsoort1 = node.innertext;    break;   case "routetekst":    string route1 = node.innertext;    break;   case "vertrekspoor":    string spoor1 = node.innertext;    break;   case "opmerkingen":    xmlnode opnode = node.firstchild;   if(opnode!=null)   {     string opmerking1 = opnode.innertext;    }   break;   }   } }