Question

In my project, I am trying to open a text file. Well the below code works but when the user click the button again and again, many files are being opened. (which I dont want)

System.Diagnostics.Process.Start(filePath);

I also tried this, File.Open and File.OpenText which are not opening the text file and also not showing any error (tried with try catch block)

File.Open(filePath); (or)
File.OpenText(filePath); (or)
FileStream fileStream = new FileStream(filePath, FileMode.Open);

I also tried this: (ERROR : Cannot be accessed with instance reference qualify with a type name instead)

System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.Start(filePath);  /*red scribbles here*/
proc.WaitForExit();

How to show only one instance of the Text file(.txt). Am I doing something wrong in my attempts? please suggest.

EDIT:

I want to open other text files afterwards but not the same and also the application should be accessible after opening a text file(or many). I have only one form.

Answer 1

Create a dictionary at the form level:

public Dictionary<string, Process> OpenedProcesses = new Dictionary<string, Process>(StringComparer.OrdinalIgnoreCase);

Now change how you open the file (note the HasExited check – this is needed so that the user can close the Notepad and reopen it):

// make sure that path is always in form C:\Folder\file.txt - less chance of different 
// paths pointing to the same file.
filePath = System.IO.Path.GetFullPath(filePath);

Process proc;
if (this.OpenedProcesses.TryGetValue(filePath, out proc) && !proc.HasExited)
{
    MessageBox.Show("The file is already open!");
    // it could be possible to activate the window of the open process but that is another question on its own.
    return;
}

proc = System.Diagnostics.Process.Start(filePath);
this.OpenedProcesses[filePath] = proc;