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How would one add a string value to a newly created xml file in c#?

In my code I'm creating a new xml file with linq to xml and I have a specific format of xml that I'm trying to put into the xml file on creation. However, when I put the string variable in it gives the error "non white space characters cannot be added to content." How would I correctly add that string value to the xml file?
string firstPart = #"<?xml version=""1.0"" encoding=""utf-8""?>
< wiidisc version = ""1"" >
< id game = ""RMCE"" disc = ""0"" version = ""0"" >
</ id > ";
XDocument doc = new XDocument(firstPart);
doc.Save(riivolutionXmls + #"\" + xmlFileName + ".xml");

This isn't valid XML, you're missing a closing tag for wiidisc.
Also I don't think you can use the constructor to create an XDocument from a string, I think you have to use the XDocument.Parse method:
https://learn.microsoft.com/en-us/dotnet/api/system.xml.linq.xdocument.parse?view=netframework-4.7.2#System_Xml_Linq_XDocument_Parse_System_String_

XmlWriter trimming my string

I am trying to return an XML string as a CLOB from Oracle stored procedure to C# string.
Then I am write this string to a file using XmlWriter class.
My code looks like following:
string myString= ((Oracle.ManagedDataAccess.Types.OracleClob)(cmd.Parameters["paramName"].Value)).Value.ToString();
string fileName = DateTime.Now.ToString("yyyyMMddHHmmss");
var stream = new MemoryStream();
var writer = XmlWriter.Create(stream);
writer.WriteRaw(myString);
stream.Position = 0;
var fileStreamResult = File(stream, "application/octet-stream", "ABCD"+fileName+".xml");
return fileStreamResult;
When I checked my CLOB output it returns completely to myString.
When I check my end result, XML file is trimmed at the end.
My string will be huge for ex: Length of 3382563 and more.
Is there any setting for XmlWriter to write the complete string to file.
Thanks in advance.

Sounds like all you want to do is grab some string value out of your Database, and write that string value in a text file. The string being xml does not actually force you into using an XML specific class or method unless you want to do XML specific operations, which I do not see in your snippet. Therefore, I suggest you simply grab the string value and spit it out in a file in the easiest way.
string myString = " blah blah blah keep my spaces ";
using (StreamWriter sw = new StreamWriter(#"M:\StackOverflowQuestionsAndAnswers\XMLWriterTrimmingString_45380476\bin\Debug\outputfile.xml"))
{
sw.Write(myString);
}

How to use replace with tricky characters in C#?

I am trying to replace within a string
<?xml version="1.0" encoding="UTF-8"?>
<response success="true">
<output><![CDATA[
And
]]></output>
</response>
with nothing.
The problem I am running into is the characters <> and " characters are interacting within the replace. Meaning, it's not reading those lines as a full string all together as one but breaking the string when it comes to a <> or ". Here is what I have but I know this isn't right:
String responseString = reader.ReadToEnd();
responseString.Replace(#"<<?xml version=""1.0"" encoding=""UTF-8""?><response success=""true""><output><![CDATA[[", "");
responseString.Replace(#"]]\></output\></response\>", "");
What would be the correct code to get the replace to see these lines as just a string?

A string will never change. The Replace method works as follows:
string x = "AAA";
string y = x.Replace("A", "B");
//x == "AAA", y == "BBB"
However, the real problem is how you handle the XML response data.
You should reconsider your approach of handling incoming XML by string replacement. Just get the CDATA content using the standard XML library. It's as easy as this:
using System.Xml.Linq;
...
XDocument doc = XDocument.Load(reader);
var responseString = doc.Descendants("output").First().Value;
The CDATA will already be removed. This tutorial will teach more about working with XML documents in C#.

Given your document structure, you could simply say something like this:
string response = #"<?xml version=""1.0"" encoding=""UTF-8""?>"
+ #"<response success=""true"">"
+ #" <output><![CDATA["
+ #"The output is some arbitrary text and it may be found here."
+ "]]></output>"
+ "</response>"
;
XmlDocument document = new XmlDocument() ;
document.LoadXml( response ) ;
bool success ;
bool.TryParse( document.DocumentElement.GetAttribute("success"), out success) ;
string content = document.DocumentElement.InnerText ;
Console.WriteLine( "The response indicated {0}." , success ? "success" : "failure" ) ;
Console.WriteLine( "response content: {0}" , content ) ;
And see the expected results on the console:
The response indicated success.
response content: The output is some arbitrary text and it may be found here.
If your XML document is a wee bit more complex, you can easily select the desired node(s) using an XPath query, thus:
string content = document.SelectSingleNode( #"/response/output" ).InnerText;

Extract XML Data from String

I am having a input stream which is generated when I upload a file(XML Type). I need the XML data at code behind. I am having the xml data in string by using
StreamReader stream = new StreamReader(Request.InputStream);
string x = stream.ReadToEnd();
It also contains the following data at the start of the string
------WebKitFormBoundary8na5dBbHc4ydfxVU
Content-Disposition: form-data; name="MyFile"; filename="Test 123.vfc"
Content-Type: application/octet-stream
at the end of the string
------WebKitFormBoundary8na5dBbHc4ydfxVU--
This data is not required for me. Please help me in getting the right XML String.

First you can remove the first three lines and last line from your string.
int n = 3;
string[] lines = str.Split(Environment.NewLine.ToCharArray()).Skip(n).ToArray();
string output = string.Join(Environment.NewLine, lines);
output = output.Remove(str.LastIndexOf(Environment.NewLine));
In your XML string if you don't have a root node then add it like following.
string xmlTxt = "<ROOT>" + xmlString + "</ROOT>";
If you have a root node skip above. For a well format XML string you can just use below code
XmlDocument xmlDocument = new XmlDocument();
xmlDocument.InnerXml = xmlTxt;

Embedding an xml document inside an xml string [duplicate]

This question already has answers here:
Best way to encode text data for XML
(13 answers)
Closed 2 years ago.
I have a web service that returns an xml string as results. The return string is in this format:
<ReturnValue>
<ErrorNumber>0
</ErrorNumber>
<Message>my message</Message>
</ReturnValue>
The data that I want to insert into the "message" tag is a serialized version of a custom object. The serialized format of that object contains xml and namespace declarations post serialization. When that gets thrown into the "message" tag of my return xml string, XmlSpy says that it's not well-formed. How should I get rid of the namespace declarations, or is there a different way to imbed a serialized object into an xml string?

Wrap the string in CDATA like so:
<![CDATA[your xml, which can be multi-line]]>
CDATA will inform a validator to treat the CDATA contents as ignored text. It's often the most expedient way to embed XML (or taggy non-XML content) as a string. You can run into problems if your embedded XML contains its own CDATA, but otherwise it's a simple fix.

Just make sure that your <Message> XML is encoded so that <, >, ", and & show up as <, >, " and &, respectively.
There are few built-in ways to encode the characters:
string message = System.Web.HttpUtility.HtmlEncode(serializedXml);
string message = System.Security.SecurityElement.Escape(serializedXml);
Using an XmlTextWriter to do the work for you
Use CDATA to wrap your XML
Also, this is probably a duplicate of:
Best way to encode text data for XML

Think of XML as a document not a string.
Create a node named "wrapper", and store the content of your file in it as a Base64 encoded string. The results will look like this.
<ReturnValue>
<ErrorNumber>0</ErrorNumber>
<Message>my message</Message>
<wrapper type="bin.base64">PD94bWwgdmVyc2lvbj0iMS4wIj8+PHhzbDpzdHlsZXNoZWV0IHZ
lcnNpb249IjEuMCIgeG1sbnM6eHNsPSJodHRwOi8vd3d3LnczLm9yZy8xOTk5L1hTTC9UcmFuc2Zvcm0
iIHhtbG5zOm1zeHNsPSJ1cm46c2NoZW1hcy1taWNyb3NvZnQtY29tOnhzbHQiPjx4c2w6b3V0cHV0IG1
ldGhvZD0ieG1sIiAvPjx4c2w6dGVtcGxhdGUgbWF0Y2g9Ii8iPjwveHNsOnRlbXBsYXRlPjwveHNsOnN
0eWxlc2hlZXQ+</wrapper>
</ReturnValue>
The following code shows how to add the wrapper, encode the content. Then it reverses the process to show that it all "works".
Using Base64 in XML has a number of other applications as well. For example embedding images, or other documents in XML content.
using System;
using System.IO;
using System.Xml;
public class t
{
static public string EncodeTo64(string toEncode) {
byte[] toEncodeAsBytes = System.Text.ASCIIEncoding.ASCII.GetBytes(toEncode);
string returnValue = System.Convert.ToBase64String(toEncodeAsBytes);
return returnValue;
}
static public string DecodeFrom64(string encodedData) {
byte[] encodedDataAsBytes = System.Convert.FromBase64String(encodedData);
string returnValue = System.Text.ASCIIEncoding.ASCII.GetString(encodedDataAsBytes);
return returnValue;
}
public static void Main() {
try {
//Create the XmlDocument.
XmlDocument doc = new XmlDocument();
doc.LoadXml( #"
<ReturnValue>
<ErrorNumber>0</ErrorNumber>
<Message>my message</Message>
</ReturnValue>
");
XmlNode nodeMessage = doc.SelectSingleNode( "/ReturnValue/Message" );
if( nodeMessage != null ) {
XmlDocument docImport = new XmlDocument();
docImport.Load( "docwithnamespace.xml" );
// create a wrapper element for the file, then import and append it after <Message>
XmlElement nodeWrapper = (XmlElement)doc.CreateElement( "wrapper" );
nodeWrapper.SetAttribute( "type", "bin.base64" );
nodeWrapper = (XmlElement)doc.ImportNode( nodeWrapper, true );
XmlNode ndImport = nodeMessage.ParentNode.AppendChild( nodeWrapper.CloneNode( true ) );
ndImport.InnerText = EncodeTo64( docImport.OuterXml );
doc.Save( "wrapperadded.xml" );
// Next, let's test un-doing the wrapping
// Re-load the "wrapped" document
XmlDocument docSaved = new XmlDocument();
docSaved.Load( "wrapperadded.xml" );
// Get the wrapped element, decode from base64 write to disk
XmlNode node = doc.SelectSingleNode( "/ReturnValue/wrapper" );
if( node != null ) {
// Load the content, and save as a new XML
XmlDocument docUnwrapped = new XmlDocument();
docUnwrapped.LoadXml( DecodeFrom64( node.InnerText ) );
docUnwrapped.Save( "unwrapped.xml" );
Console.WriteLine( "Eureka" );
}
}
} catch( Exception e ) {
Console.WriteLine(e.Message);
}
}
}