Transfer the local name of a XML XElement to string

Question

Hy at all, i am new in C# programming so maybe this is a easy question. I try to read an XML-file with XDocument and write out the local name of a Element as string.

For the output i use the listing window of SiemensNX but every other output window or txt-file is suitable.

Here is the Input-XML:

<?xml version="1.0" encoding="utf-8"?>

<Rootlvl>
    <Lvl_1>
        <Lvl_2/>
    </Lvl_1>
</Rootlvl>

Here is the C# code:

using System.Xml.Linq;
using NXOpen;

namespace XmlElementName
{
    class Program
    {
        private static ListingWindow lw = s.ListingWindow;


        public static void Main()
        {
            string XmlFilePath = @"C:\Users\XXX\Desktop\TestXML.xml"; //XML path
            string testnode = "Lvl_2"; //local name of a optional XML element
            lw.Open(); //open NX listinwindow for output

            //=============LoadXmlFile================
            //get main Input
            XDocument xml = XDocument.Load(XmlFilePath); //load XmlFile



            //====================WriteOutElementName====================
            XElement node;
            if (testnode == null) //if no optional Element name --> take root element of XML
            {
                node = xml.Root;
            }
            else
            {
                // Find node to passed string "testnode" --> here "Lvl_2"
                node = xml.Element(testnode);
            }

            lw.WriteLine("Test"); //Test if output works --> !yes it works

            if(node != null)
            {
                string output = node.Name.LocalName;
                //local name of XElement-variable "node" to string

                lw.WriteLine(output); //output the local name of variable "node"
            }
            else
            {
               lw.WriteLine("Element with Name = " + testnode + "not found")
            }


        }
    }
}

The Output if variable testnode = "Lvl_2" should be:

Test
Lvl_2

The Output if variable testnode = null schould be:

Test
Rootlvl

The VS Debugger show me that

node = xml.Element(testnode); //testnode = Lvl_2

can't find an element in the XML with the the name "Lvl_2". So it set "node" to "Null" and throw the following exeption:

"System.NullReferenceException: Object reference not set to an instance of an object."

But i know that "Lvl_2" is a subelement of the XML. What can i do to found this element "Lvl_2" in the XML?

What should i change in this line

node = xml.Element(testnode)

to find the element by name?

Thanks for your help guys.

Does this answer your question? [What is a NullReferenceException, and how do I fix it?](https://stackoverflow.com/questions/4660142/what-is-a-nullreferenceexception-and-how-do-i-fix-it) — Sinatr, Jan 15 '20 at 13:09
string results = (string)xml.Descendents(testnode).FirstOrDefault(); — jdweng, Jan 15 '20 at 14:28

score 0 · Answer 1 · answered Jan 15 '20 at 13:31

0

If you put a testnode that is not present in the xml then node is null and it causes an exception, do the check before printing

if (node!=null){
        string output = node.Name.LocalName;
        //local name of XElement-variable "node" to string

        lw.WriteLine(output); //output the local name of variable "node"
}

answered Jan 15 '20 at 13:31

Albondi

1,141
1
8
19

Thanks for your answere. I add your solution but it don't answere the whole question. I edit the the question to specify it. – MarcelM Jan 15 '20 at 14:15

Giorgos Betsos · Accepted Answer · 2020-01-15T14:50:03.137

0

According to MSDN:

Method XDocument.Element(XName) gets the first (in document order) child element with the specified XName.

So, in your case,

node = xml.Element(testnode)

returns null if testnode = "Lvl_2", because Lvl_2 is not a child element of the xml document (only the root node Rootlvl is considered as a child element is this context).

Try using Descendants method instead:

node = xml.Descendants(testnode).FirstOrDefault();

edited Jan 15 '20 at 14:50

answered Jan 15 '20 at 14:43

Giorgos Betsos

71,379
9
63
98

Yeah Thank you, I found the same solution with the approach of jdweng. But your answer explains very well why my first approach was wrong. – MarcelM Jan 15 '20 at 15:05

score -1 · Answer 3 · answered Jan 15 '20 at 15:01

-1

Thanks @jdweng,

i take the following changes. Now it works.

using system.linq;

and replace

node.xml.Element(testnode);

with

node = xml.Descendents(testnode).FirstOrDefault();

answered Jan 15 '20 at 15:01

MarcelM

29
4

Transfer the local name of a XML XElement to string

3 Answers3