how to write a element in a xml file

Question

I'm writing a xml file, but when i started to write elements, it comes with some errors, and when i try to read the xml fil by code, it can't find these elements.

<?xml version="1.0" encoding="utf-8" ?>
<options>
  <difficulty>
    <type name="Easy" health="6" active="0"/>
    <type name="Normal" health="4" active="1"/>
    <type name="Hard" health="2" active="0"/>
  </difficulty>
  <soundvolume>
    <type name="Sound" value="100"/>
    <type name="tempSound" value="100"/>
  </soundvolume>
</options>

this is the xml code so far, but i don't want to continue if i can't get it to work.

this is the errors I get:

Could not find schema information for the element 'options'.

and every element has the same error. I use visual studio 2013 and have a Windows Forms Application C# project

This is how i read the xml file:

StreamReader sr = new StreamReader("Options.xml");
            String xmlsr = sr.ReadToEnd();
            sr.Close();

            XElement xDocumentSr = XElement.Parse(xmlsr);
            XElement xOptionsSr = xDocumentSr.Element("options");
            XElement xDifficultySr = xOptionsSr.Element("difficulty");

            foreach (XElement xType in xDifficultySr.Descendants("type"))
            {
                if(Convert.ToInt32(xType.Attribute("activate").Value) == 1)
                {
                    labDifficulty.Text = xType.Attribute("name").Value;
                }
            }

the error happens when her:

XElement xOptionsSr = xDocumentSr.Element("options");

and i get this error:

An unhandled exception of type 'System.NullReferenceException' occurred in Splash Screen.exe

Additional information: Object reference not set to an instance of an object.

and when in debug mode i can see that the element is = null

Answer 1

Here:

XElement xDocumentSr = XElement.Parse(xmlsr);
XElement xOptionsSr = xDocumentSr.Element("options");

The xDocumentSr itself options.So you are looking for options element inside of itself.You don't need that xDocumentSr.Element("options");.And you can simplify your code like this:

var xmlDocument = XDocument.Load("Options.xml");

var element = xmlDocument.Root
              .Element("difficulty")
              .Elements("type")
              .FirstOrDefault(x => (int)x.Attribute("active") == 1);

if(element != null)
      labDifficulty.Text = element.Attribute("name").Value;