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I used xmlDocument :

C#
XmlDocument xml = new XmlDocument();
                           xml.LoadXml(text); // suppose that myXmlString contains "<Names>...</Names>"

                           XmlNodeList xnList = xml.SelectNodes("Etichetta");
                           foreach (XmlNode xn in xnList)
                           {
                               string firstName = xn["Barcode"].InnerText;
                               string lastName = xn["Materiale"].InnerText;

                           }




but when I run code I have this error: An unhandled exception of type 'System.Xml.XmlException' occurred in System.Xml.dll

What I have tried:

I tried to use XDocument but I don't know I can write to find my root, my file xml is
Posted
Updated 11-Apr-23 9:39am
v3
Comments
0x01AA 5-Apr-23 9:56am    
1. XmlNodeList xnList = xml.SelectNodes("Etichetta");
should be XmlNodeList xnList = xml.SelectNodes("PrintDymo/Etichetta");
2. Barcode and Materiale are Attributes of _individual_ Campi nodes, which at least for me makes not big sense.
PIEBALDconsult 5-Apr-23 10:03am    
or xml.SelectNodes("//Etichetta")
or xml.SelectNodes("*/Etichetta") (maybe?)
0x01AA 5-Apr-23 10:13am    
Thanks for this. I'm not experienced with XPath, that's why I did not post a solution but only a comment. At least I tested it before posting ;)
Member 14594285 5-Apr-23 10:16am    
I don't understand, I have always the same error
PIEBALDconsult 5-Apr-23 10:18am    
Yes, well, I may be able to assist in a bit.
Do you have no control over the format of the XML file?

Try this code, which will list the values of the Campi attributes.
C#
XmlDocument xml = new XmlDocument();
XmlTextReader reader = new XmlTextReader("Filename.xml");
xml.Load(reader);

XmlNodeList xnList = xml.SelectNodes("PrintDymo/Etichetta");
foreach (XmlNode xn in xnList)
{
    XmlNodeList campiList = xn.SelectNodes("Campi");
    foreach (XmlNode cn in campiList)
    {
        XmlAttributeCollection attribs = cn.Attributes;
        Console.WriteLine($"attribute: {attribs[0].Value}");
    }
}

The Campi elements do not contain anything except a single attribute, which as others have pointed out, does not make a lot of sense. An element should contain all valid attrivbutes together, something like:
XML
<Campi Barcode ="123456789012" Data ="05/04/2023" Materiale="Pellame"/>


[edit]
Changed to SelectNodes.
[/edit]

[edit]
With Campi entries correctly formatted try:
C#
XmlNodeList xnList = xml.SelectNodes("PrintDymo/Etichetta/Campi");;
foreach (XmlNode xn in xnList)
{
    string firstName = xn.Attributes["Barcode"].InnerText;
    string lastName = xn.Attributes["Materiale"].InnerText;
    Console.WriteLine($"firstName: {firstName}, lastName: {lastName}");
}


[/edit]
 
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v3
Comments
PIEBALDconsult 5-Apr-23 10:26am    
Ew :: makes the sign of the Cross at "GetElementsByTagName" ::
Richard MacCutchan 5-Apr-23 10:28am    
Yes, I know, but I am not an XML expert like you. I mainly wanted to demonstrate why the code and the XML are not a good match.
Richard MacCutchan 5-Apr-23 10:40am    
Better?
PIEBALDconsult 5-Apr-23 10:46am    
Sigh of relief.
Member 14594285 5-Apr-23 11:21am    
thanks
Hi
Here is a solution by using LINQ to xml

C#
XDocument doc = XDocument.Load("YourData.xml");

var CampiList = (from e in doc.Descendants("Etichetta")
                 let c = e.Elements("Campi")
                 select new
                 {
                     Barcode = c.Attributes("Barcode").First().Value,
                     Data = c.Attributes("Data").First().Value,
                     Materiale = c.Attributes("Materiale").First().Value
                 }).ToList();

foreach(var campi in CampiList)
{
    Console.WriteLine($"firstname {campi.Barcode} lastname {campi.Materiale}.");
}
 
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v6

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