XDocument.ToString() drops XML Encoding Tag

Solution 1:

Either explicitly write out the declaration, or use a StringWriter and call Save():

using System;
using System.IO;
using System.Text;
using System.Xml.Linq;

class Test
{
    static void Main()
    {
        string xml = @"<?xml version='1.0' encoding='utf-8'?>
<Cooperations>
  <Cooperation />
</Cooperations>";

        XDocument doc = XDocument.Parse(xml);
        StringBuilder builder = new StringBuilder();
        using (TextWriter writer = new StringWriter(builder))
        {
            doc.Save(writer);
        }
        Console.WriteLine(builder);
    }
}

You could easily add that as an extension method:

public static string ToStringWithDeclaration(this XDocument doc)
{
    if (doc == null)
    {
        throw new ArgumentNullException("doc");
    }
    StringBuilder builder = new StringBuilder();
    using (TextWriter writer = new StringWriter(builder))
    {
        doc.Save(writer);
    }
    return builder.ToString();
}

This has the advantage that it won't go bang if there isn't a declaration :)

Then you can use:

string x = doc.ToStringWithDeclaration();

Note that that will use utf-16 as the encoding, because that's the implicit encoding in StringWriter. You can influence that yourself though by creating a subclass of StringWriter, e.g. to always use UTF-8.

Solution 2:

The Declaration property will contain the XML declaration. To get the contents plus declaration, you can do the following:

tb_output.Text = xml.Declaration.ToString() + xml.ToString()

Solution 3:

use this:

output.Text = String.Concat(xml.Declaration.ToString() , xml.ToString())

Solution 4:

I did like this

        string distributorInfo = string.Empty;

        XDocument distributors = new XDocument();

     //below is important else distributors.Declaration.ToString() throws null exception
        distributors.Declaration = new XDeclaration("1.0", "utf-8", "yes"); 

        XElement rootElement = new XElement("Distributors");
        XElement distributor = null;
        XAttribute id = null;

        distributor = new XElement("Distributor");
        id = new XAttribute("Id", "12345678");
        distributor.Add(id);
        rootElement.Add(distributor);

        distributor = new XElement("Distributor");
        id = new XAttribute("Id", "22222222");

        distributor.Add(id);

        rootElement.Add(distributor);         

        distributors.Add(rootElement);

        distributorInfo = String.Concat(distributors.Declaration.ToString(), distributors.ToString());

Please see below for what I get in distributorInfo

<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<Distributors>
  <Distributor Id="12345678" />
  <Distributor Id="22222222" />
  <Distributor Id="11111111" />
</Distributors>

Solution 5:

Similar to the other +1 answers, but a bit more detail about the declaration, and a slightly more accurate concatenation.

<xml /> declaration should be on its own line in a formatted XML, so I'm making sure we have the newline added. NOTE: using Environment.Newline so it will produce the platform specific newline

// Parse xml declaration menthod
XDocument document1 =
  XDocument.Parse(@"<?xml version=""1.0"" encoding=""iso-8859-1""?><rss version=""2.0""></rss>");
string result1 =
  document1.Declaration.ToString() +
  Environment.NewLine +
  document1.ToString() ;

// Declare xml declaration method
XDocument document2 = 
  XDocument.Parse(@"<rss version=""2.0""></rss>");
document2.Declaration =
  new XDeclaration("1.0", "iso-8859-1", null);
string result2 =
  document2.Declaration.ToString() +
  Environment.NewLine +
  document2.ToString() ;

Both results produce:

<?xml version="1.0" encoding="iso-8859-1"?>
<rss version="2.0"></rss>