Find root of a file XML

Question

0.00/5 (No votes)

See more:

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 5-Apr-23 3:07am

Member 14594285

Updated 11-Apr-23 9:39am

v3

Add a Solution

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?

PIEBALDconsult 5-Apr-23 10:17am

XPath and Regular Expressions are my bread and butter. :D

0x01AA 5-Apr-23 10:22am

That sounds good (for OP), please take over :D

Member 14594285 5-Apr-23 10:13am

so:
XmlDocument xml = new XmlDocument();
xml.LoadXml(text); // suppose that myXmlString contains "<names>..."

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

}

Member 15627495 5-Apr-23 10:21am

'Xelement' are good tools for 'Xml', as 'Xdocument' are complete 'Xml tree', the Xelement is flexible.

Richard Deeming 5-Apr-23 10:24am

An unhandled exception of type 'System.Xml.XmlException' occurred in System.Xml.dll

Passing the XML you've posted to the code you've posted does not produce that error. You need to debug your code to see what the actual text you're trying to parse contains, and look at the full details of the exception.

2 solutions

Solution 2

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}.");
}

Posted 11-Apr-23 9:39am

Jean Ferre

Updated 14-Apr-23 6:49am

v6

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Richard MacCutchan · Accepted Answer · 2023-04-05T04:23:00

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]