How I can read three different values in a XML

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Hello, I have the follwing xml file:

<Directorys>
  <DIR>
    <name>Installation Path</name>
    <path>C:\Users\briganti\Documents\Lavoro\PIPPO</path>
  </DIR>
  <DIR>
    <name>Temp directory</name>
    <path>C:\Users\briganti\Documents\Lavoro\PLUTO</path>
  </DIR>
  <DIR>
    <name>Test List Path</name>
    <path>C:\Users\briganti\Documents\Lavoro\PAPERINO</path>
  </DIR>
</Directorys>

and I want to read and store the three path value in three different string variables using C# language. I used the following code and it works but is it possible to do the same thing more simply?

What I have tried:

XmlDocument doc = new XmlDocument();
            doc.Load(Directory.GetCurrentDirectory() + @"\programconfig.xml");
            XmlNodeList paths = doc.DocumentElement.SelectNodes("/Directorys/DIR/path");
            var myList = new List<string>();
            foreach (XmlNode path in paths)
            {
                // Create list
                myList.Add(path.InnerText);
                //MessageBox.Show(path.InnerText);

            }
            // Convert to array
            var myArray = myList.ToArray();
            t_Installazione.Text = myArray[0];
            t_Temp.Text = myArray[1];
            t_testlist.Text = myArray[2];

Posted 21-Apr-20 3:16am

Jimdigriz76

Updated 21-Apr-20 8:23am

v2

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CHill60 21-Apr-20 9:32am

You need to show us the code you have tried and tell us what the problem is

3 solutions

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Garth J Lancaster · Answer 1 · 2020-04-21T03:48:00

I'm feeling generous - must be an ISO/Alcohol thing .... this might start you off

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

foreach(XmlNode row in doc.SelectNodes("//DIR"))
{
   string rowName = row.SelectSingleNode("//path").InnerText;
   Console.WriteLine("{0}", rowName);
}

Another approach might be to deserialize the XML to a list of Objects, eg with name & path attributes Convert XML to C# Object[^]

RickZeeland · Answer 2 · 2020-04-21T06:55:00

Solution 2

You can also use XElement, which is a bit lighter than using XmlDocument:

//using System.Linq;
//using System.Xml.Linq;

XElement x = XElement.Load("programconfig.xml");

foreach (var item in x.Elements())
{
    Debug.Print(item.Element("path").Value.ToString());
}

Posted 21-Apr-20 6:55am

RickZeeland

Richard Deeming · Answer 3 · 2020-04-21T08:24:00

You might want to consider what happens if the DIR elements are not in the expected order, or any of the paths are missing.

C#

string filePath = Path.GetFullPath(@".\programconfig.xml");
XElement doc = XElement.Load(filePath);

Dictionary<string, string> paths = doc.Descendants("DIR").ToDictionary(
    d => (string)d.Element("name"), 
    d => (string)d.Element("path"), 
    StringComparer.OrdinalIgnoreCase);

string value;
if (paths.TryGetValue("Installation Path", out value))
{
    t_Installazione.Text = value;
}
if (paths.TryGetValue("Temp directory", out value))
{
    t_Temp.Text = value;
}
if (paths.TryGetValue("Test List Path", out value))
{
    t_testlist.Text = value;
}

How I can read three different values in a XML

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