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C#. Writing settings to XML file

I have some names of files. I am creating a XML file to store them. When application runs second time it should recognize those file names. Here is my code:
XmlWriter writer = XmlWriter.Create("settings.xml");
writer.WriteStartElement("DialogCreater");
writer.WriteElementString("conditionsFile", conditionsFile);
writer.WriteEndElement();
writer.Close();
But when I use those file names like this:
string[] lines = File.ReadAllLines(charactersFile);
It gives me an error.
System.NotSupportedException
Here how I read them from XML file:
XmlDocument doc = new XmlDocument();
doc.Load("settings.xml");
XmlNodeList node = doc.GetElementsByTagName("files");
foreach (XmlNode n in node[0].ChildNodes)
{
if (n.Name == "conditionsFile") conditionsFile = n.InnerText;
}

NotSupportedException
path is in an invalid format according to msdn documentation. you can fix it by checking path exist or not before reading from itMore info
string path = #"c:\temp\MyTest.txt";
// This text is added only once to the file.
if (!File.Exists(path))
{
//file exist
}

Appending inside a parent tag using XmlWriter

I am using this code to write to a xml document. Issue is every time I call this code it overrides previously written Tags element.
However I want to append multiple Tag elements inside the Tags elements. How can I make such sets using XmlWriter?
using (XmlWriter writer = XmlWriter.Create(path))
{
writer.WriteStartElement("Tags");
writer.WriteElementString("Tag", tagName);
writer.WriteEndElement();
}
I found out over the net few solution involving LINQ, with which I am not very good at. So I am looking something without it?

This can be done via Linq Xml by:
public void AddTagToXml(string path, string tag)
{
XDocument doc;
// Load the existing file
if (File.Exists(path)) doc = XDocument.Load(path);
else
{
// Create a new file with the parent node
doc = new XDocument(new XElement("Tags"));
}
doc.Root.Add(new XElement("tag", tag));
doc.Save(path);
}
It's not terribly efficient as the file is opened and saved on each function call, but it covers your requirements.

Trying to Load an XML File Upon Form Load, Getting Error

In C# am trying to check to see if an XML file is created, if not create the file and then create the xml declaration, a comment and a parent node.
When I try to load it, it gives me this error:
"The process cannot access the file 'C:\FileMoveResults\Applications.xml' because it is being used by another process."
I checked the task manager to ensure it wasn't open and sure enough there were no open applications of it. Any ideas of what's going on?
Here is the code I am using:
//check for the xml file
if (!File.Exists(GlobalVars.strXMLPath))
{
//create the xml file
File.Create(GlobalVars.strXMLPath);
//create the structure
XmlDocument doc = new XmlDocument();
doc.Load(GlobalVars.strXMLPath);
//create the xml declaration
XmlDeclaration xdec = doc.CreateXmlDeclaration("1.0", null, null);
//create the comment
XmlComment xcom = doc.CreateComment("This file contains all the apps, versions, source and destination paths.");
//create the application parent node
XmlNode newApp = doc.CreateElement("applications");
//save
doc.Save(GlobalVars.strXMLPath);
Here is the code I ended up using to fix this issue:
//check for the xml file
if (!File.Exists(GlobalVars.strXMLPath))
{
using (XmlWriter xWriter = XmlWriter.Create(GlobalVars.strXMLPath))
{
xWriter.WriteStartDocument();
xWriter.WriteComment("This file contains all the apps, versions, source and destination paths.");
xWriter.WriteStartElement("application");
xWriter.WriteFullEndElement();
xWriter.WriteEndDocument();
}

File.Create() returns a FileStream that locks the file until it's closed.
You don't need to call File.Create() at all; doc.Save() will create or overwrite the file.

I would suggest something like this:
string filePath = "C:/myFilePath";
XmlDocument doc = new XmlDocument();
if (System.IO.File.Exists(filePath))
{
doc.Load(filePath);
}
else
{
using (XmlWriter xWriter = XmlWriter.Create(filePath))
{
xWriter.WriteStartDocument();
xWriter.WriteStartElement("Element Name");
xWriter.WriteEndElement();
xWriter.WriteEndDocument();
}
//OR
XmlDeclaration xdec = doc.CreateXmlDeclaration("1.0", null, null);
XmlComment xcom = doc.CreateComment("This file contains all the apps, versions, source and destination paths.");
XmlNode newApp = doc.CreateElement("applications");
XmlNode newApp = doc.CreateElement("applications1");
XmlNode newApp = doc.CreateElement("applications2");
doc.Save(filePath); //save a copy
}
The reason your code is currently having problems is because of: File.Create creates the file and opens the stream to the file, and then you never make use of it (never close it) on this line:
//create the xml file
File.Create(GlobalVars.strXMLPath);
if you did something like
//create the xml file
using(Stream fStream = File.Create(GlobalVars.strXMLPath)) { }
Then you would not get that in use exception.
As a side note XmlDocument.Load will not create a file, only work with an already create one

You could create a stream, setting the FileMode to FileMode.Create and then use the stream to save the Xml to the path specified.
using (System.IO.Stream stream = new System.IO.FileStream(GlobalVars.strXMLPath, FileMode.Create))
{
XmlDocument doc = new XmlDocument();
...
doc.Save(stream);
}

How to read XML document and display the output as string in c#

consider my source file looks like this.
<Content xmlns="uuid:4522eb85-0a47-45f9-8e2b-1x82c78xx920">
<first>Hello World.This is Fisrt field</first>
<second>Hello World.This is second field</second>
</Content>
I want to write a code, which read this xml document from a location and display it as string.
say name of the xml file is helloworld.xml.
Location: D:\abcd\cdef\all\helloworld.xml.
I have tried the following, but i was unable to do it.
XmlDocument contentxml = new XmlDocument();
contentxml.LoadXml(#"D:\abcd\cdef\all\helloworld.xml");
Response.Write("<BR>" + contentxml.ToString());
Response.write is displaying nothing. Correct me if i missed any thing. Its not creating any component and error is coming.
I have also tried this,
XmlDocument contentxml = new XmlDocument();
try
{
contentxml.LoadXml(#"D:\abcd\cdef\all\helloworld.xml");
}
catch (XmlException exp)
{
Console.WriteLine(exp.Message);
}
StringWriter sw = new StringWriter();
XmlTextWriter xw = new XmlTextWriter(sw);
contentxml.WriteTo(xw);
Response.Write("<BR>" + sw.ToString());
But i did not find the any output.
I want to read a XML file from a location and display it as it is as string.
Can anyone help on this.
Thank you,
Muzimil.

You need the OuterXml property:
Response.Write("<BR>" + contentxml.OuterXml);
Also you are loading a file not xml so use
contentxml.Load(#"D:\abcd\cdef\all\helloworld.xml");
instead of
contentxml.LoadXml(#"D:\abcd\cdef\all\helloworld.xml");

Do you really have to deserialize the XML at all? Why not just read it as a text file? Something like..
String text = File.ReadAllText(#"D:\abcd\cdef\all\helloworld.xml");
Response.Write(text);
With appropriate error handling obviously..

I would try using the XDocument class:
//load the document from file
var doc = XDocument.Load("..."); //== path to the file
//write the xml to the screen
Response.Write(doc.ToString());
If you want to use an XmlDocument instead, you would want to use Load instead LoadXml.

If you want to simply write a file to the output, you can do Response.WriteFile.

try this
XmlTextReader reader = new XmlTextReader (#"D:\abcd\cdef\all\helloworld.xml");
while (reader.Read())
{
Console.WriteLine(reader.Name);
}
Console.ReadLine();

String text = File.ReadAllText(Server.MapPath("~/App_Data/sample.xml"));
txtData.Text = text;

How to Update a XML Node?

It is easy to read an XML file and get the exact Node Text, but how do I Update that Node with a new value?
To read:
public static String GetSettings(SettingsType type, SectionType section)
{
XmlTextReader reader = new XmlTextReader(HttpContext.Current.Request.MapPath(APPSETTINGSPATH));
XmlDocument document = new XmlDocument();
document.Load(reader);
XmlNode node = document.SelectSingleNode(
String.Format("/MyRootName/MySubNode/{0}/{1}",
Enum.Parse(typeof(SettingsType), type.ToString()),
Enum.Parse(typeof(SectionType), section.ToString())));
return node.InnerText;
}
to write ...?
public static void SetSettings(SettingsType type, SectionType section, String value)
{
try
{
XmlTextReader reader = new XmlTextReader(HttpContext.Current.Request.MapPath(APPSETTINGSPATH));
XmlDocument document = new XmlDocument();
document.Load(reader);
XmlNode node = document.SelectSingleNode(
String.Format("/MyRootName/MySubNode/{0}/{1}",
Enum.Parse(typeof(SettingsType), type.ToString()),
Enum.Parse(typeof(SectionType), section.ToString())));
node.InnerText = value;
node.Update();
}
catch (Exception ex)
{
throw new Exception("Error:", ex);
}
}
Note the line, node.Update(); does not exist, but that's what I wanted :)
I saw the XmlTextWriter object, but it will write the entire XML to a new file, and I just need to update one value in the original Node, I can save as a new file and then rename the new file into the original name but... it has to be simpler to do this right?
Any of you guys have a sample code on about to do this?
Thank you

You don't need an "update" method - setting the InnerText property updates it. However, it only applies the update in memory. You do need to rewrite the whole file though - you can't just update a small part of it (at least, not without a lot of work and no out-of-the-box support).

XmlDocument.Load has an overload that will take the filename directly so there is no need for the reader.
Similarly when you are done XmlDocument.Save will take a filename to which it will save the document.

The nodeValue property can be used to change the value of a text node.
The following code changes the text node value of the first element:
Example:
xmlDoc=loadXMLDoc("books.xml");
x=xmlDoc.getElementsByTagName("title")[0].childNodes[0];
x.nodeValue="Easy Cooking";
source: http://www.w3schools.com/DOM/dom_nodes_set.asp

You're updating the node in an in-memory representation of the xml document, AFAIK there's no way to update the node directly in the physical file. You have to dump it all back to a file.