Starting Notepad with a Known Filename within C#

Try:

Process p = new Process();
ProcessStartInfo psi = new ProcessStartInfo("Notepad.Exe", @"D:\Temp\MyLargeTextFile.txt");
p.StartInfo = psi;
p.Start();

Consider using the shell execute command. In this case, the associated application for txt-files will open. In many cases this will be standard windows notepad, but some people, like me ;-), prefer different editors like Notepad++ etc.

Process foo = new Process();
foo.StartInfo.UseShellExecute = true;
foo.StartInfo.FileName = "filename.txt";
foo.Start();

Use the advice of Mr Wed

Ok use the following code here............

System.Diagnostics.ProcessStartInfo pi = new System.Diagnostics.ProcessStartInfo("notepad.exe", "D:\\reg.txt");
System.Diagnostics.Process.Start(pi);

Hope this helps.........

using System.IO;
using System.Diagnostics;

public void OpenNotePadPlusPlus()
{
string filename = @"C:\Users\Brian\Documents\file I wanna open.txt");
FileInfo fi = new FileInfo(filename);
if (fi.Exists == false) return;
string fname = "\u0022"+filename+"\u0022"; // unicode for double quote marks
string editorpath =
"\u0022"+@"C:\Program Files (x86)\Notepad++\Notepad++.exe"+"\u0022";
Process process;
try
{
process = Process.Start(editorpath,fname);
}
catch (Exception ee)
{
string msg = ee.Message;
// do something with message
return;
}
process.WaitForExit();
// I needed the above wait so i could reprocess stuff after notepad++ edits
}