Opening any file

[Dylan van Heerden]

How do you open a file for example an XML file in coding? I tried "File.Open(Path including filename,FileMode.Open,FileAccess.Read)" but it does not seem to work. Am I missing something? Sweet

[the last free name]

I usually go for something like this:- XmlTextReader reader = new XmlTextReader(file) and then read it with:- while (!reader.EOF) { if(reader.NodeType == XmlNodeType.Element) { switch (reader.Name) { case "AnInterestingNode": // do stuff break; } } reader.Read() // advances the reader } The only thing to look out for is if you use reader.ReadOuterXML(); in a case statemant as this advances the reader, and the 'reader.Read();' call means you will miss a node. (This very thing has me scratching my head for about an hour!)

[Dylan van Heerden]

Ok what I want to do is open the file like you would normally double click on it in a directory. Is it possible? Sweet

[the last free name]

you mean open a file in a supporting application eg Notepad and a .txt file? if so Process.Start("Notepad.exe", "C:\\temp.txt"); If environment variable 'path' doesn't complete the app path you will have to give it in full.

[sreejith ss nair]

So i think you know how to get a required file and file name. //pass your file name as parameter //use this code below to open your file like normal double clicking. System.Diagnostics.Process.Start("filename") :-O Sreejith S S Nair

[Dylan van Heerden]

I tried the following try { DirectoryInfo dirInfo = Directory.GetParent("C:\\Club"); FileInfo[] Files = dirInfo.GetFiles("*.xml"); MessageBox.Show(Files[0].ToString()); Process.Start(Files[0].ToString()); } catch(Exception Error) { MessageBox.Show(Error.Message); } and I recieve the following error "The system cannot find the file specified" :doh: Sweet

[Dave Kreskowiak]

_Hacker wrote: MessageBox.Show(Files[0].ToString()); Process.Start(Files[0].ToString()); I bet the MessageBox didn't show the FULL path to the file you tried to open, did it? The following code changes will work because the code will no longer assume a default directory with using just Files[]:

try
{
String parentPath = Directory.GetParent("C:\\Club");
String fullFilePath;
DirectoryInfo dirInfo = New DirectoryInfo(parentPath);
 
FileInfo[] Files = dirInfo.GetFiles("*.xml");
fullFilePath = Path.Combine(parentPath, Files[0]);
MessageBox.Show(fullFilePath);
Process.Start(fullFilePath);
}
catch(Exception Error)
{
MessageBox.Show(Error.Message);
}

RageInTheMachine9532 "...a pungent, gastly, stinky piece of cheese!" -- The Roaming Gnome

[Dylan van Heerden]

Thank you all... It works! Sweet