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How can i parse simple xml in .net

Posted on 2010-08-17
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Last Modified: 2012-05-10
I have an xml string from some third party software. It contains metadata about some object and is delivered to me as string similar to sXml below.

//---------------------------------------
String sXml = "<Latitude>69.3</Latitude><Longitude>16.12</Longitude><Altitude>23</Altitude><Weather>Feels like winter</Weather>";

Now I want to extract this data and put it into my variables.

double Lat = -1 ;
double Lon = -1;
int Alt = -1;
String Weather = "" ;

DoTheFunkyThing(sXml, ref Lat, ref Lon, ref Alt, ref Weather); //Equivalent to magic

//---------------------------------------


Can someone help me figure out how to perform the magic? I'm able to hardcode a solution for this exact sXml, however I would like to learn how to use the classes in System.Xml namespace.

The way I see it, XML couldn't be any easier than my string; So why is it so hard to solve?

Arild


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Question by:arildj78
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7 Comments
 
LVL 18

Expert Comment

by:Anil Golamari
ID: 33455970
using System;
using System.Xml;

namespace ReadXMLfromFile
{
    /// <summary>
    /// description for Class1.
    /// </summary>
    class Class1
    {
        static void Main(string[] args)
        {
            XmlTextReader reader = new XmlTextReader ("sXml .xml");
            while (reader.Read())
            {
                switch (reader.NodeType)
                {
                    case XmlNodeType.Element: // The node is an element.
                        Console.Write("<" + reader.Name);
                        Console.WriteLine(">");
                        break;
                    case XmlNodeType.Text: //Display the text in each element.
                        Console.WriteLine (reader.Value);
                        break;
                    case XmlNodeType.EndElement: //Display the end of the element.
                        Console.Write("</" + reader.Name);
                        Console.WriteLine(">");
                        break;
                }
            }
            Console.ReadLine();
        }
    }
}
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LVL 18

Expert Comment

by:Anil Golamari
ID: 33455973
0
 
LVL 16

Accepted Solution

by:
kris_per earned 260 total points
ID: 33456102

Below is the DoTheFunkyThing method:
static void Main()
        {
            String sXml = "<Latitude>69.3</Latitude><Longitude>16.12</Longitude><Altitude>23</Altitude><Weather>Feels like winter</Weather>";
            double lat=0, lon=0, alt=0;
            string weather = "";

            DoTheFunkyThing(sXml, ref lat, ref lon, ref alt, ref weather);
}

public static void DoTheFunkyThing(string sXml, ref double Lat, ref double Lon, ref double Alt, ref string Weather)
        {
            sXml = string.Format("<mydata>{0}</mydata>", sXml);

            XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.LoadXml(sXml);

            string val = GetNodeValue(xmlDoc.DocumentElement, "Latitude");
            Lat = Convert.ToDouble(val);

            val = GetNodeValue(xmlDoc.DocumentElement, "Longitude");
            Lon = Convert.ToDouble(val);

            val = GetNodeValue(xmlDoc.DocumentElement, "Altitude");
            Alt = Convert.ToDouble(val);

            Weather = GetNodeValue(xmlDoc.DocumentElement, "Weather");
        }

        private static string GetNodeValue(XmlNode parentNode, string nodeName)
        {
            XmlNode node = parentNode.SelectSingleNode(nodeName);
            return node.InnerText;
        }

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LVL 51

Assisted Solution

by:Huseyin KAHRAMAN
Huseyin KAHRAMAN earned 240 total points
ID: 33456104
try this
        String sXml = "<data><Latitude>69.3</Latitude><Longitude>16.12</Longitude><Altitude>23</Altitude><Weather>Feels like winter</Weather></data>";

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

        double Lat = double.Parse(doc.SelectSingleNode("/data/Latitude").InnerText);
        double Lon = double.Parse(doc.SelectSingleNode("/data/Longitude").InnerText);
        int Alt = int.Parse(doc.SelectSingleNode("/data/Altitude").InnerText);

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LVL 2

Author Comment

by:arildj78
ID: 33459698
Are there any chance of extracting the value by entering the tag as an argument?
Something similar to:

String sXml = "<Latitude>69.3</Latitude><Longitude>16.12</Longitude>" +
                      "<Altitude>23</Altitude><Weather>Feels like winter</Weather>";

String Weather = ExtractElementByTag( "Weather" );
0
 
LVL 2

Author Comment

by:arildj78
ID: 33459863
Sorry... forgot to refresh (for 6 hours....)

kris_per and HainKurt: Pretty much the same solution with the same timestamp. Did one of you cheat?
Anyway, your solutions was exactly what I was looking for. I have no idea why it was so hard to find that with google and msdn.

Below is the first part of my own implementation. Works perfectly :)
Lucky85: I'm sorry, but your solution was the same that I got from msdn and google; a sequential read through the document. Thanks for your time.


sXml = string.Format("<data>{0}</data>", PolyData0.ExtraData);

System.Xml.XmlDocument xmlDoc = new System.Xml.XmlDocument();
xmlDoc.LoadXml(sXml);

string val = xmlDoc.SelectSingleNode("/data/Lat").InnerText;

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LVL 2

Author Closing Comment

by:arildj78
ID: 33459886
kris_per did supply the exact method I was asking for.
HainKurt did supply same solution, but with slightly shorter code (easy to read though).

CHEERS :)
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