The system cannot find the file specified Exception in Process Start

I am having some exception in starting a process from C# Following is the cod开发者_StackOverflow社区e

Process myProcess = new Process();  
try  
{  
    myProcess.StartInfo.UseShellExecute = true;  
    myProcess.StartInfo.FileName = "c:\\windows\\system32\\notepad.exe C:\\Users\\Karthick\\AppData\\Local\\Temp\\5aau1orm.txt";  
    myProcess.StartInfo.CreateNoWindow = false;  
    myProcess.Start();  
}  
catch (Exception e)  
{  
    Console.WriteLine(e.Message);  
}

And sometimes I get the exception "The filename, directory name, or volume label syntax is incorrect" if useShellExecute is set to false

Any Ideas as to why this is not coming out correctly

You can't put an entire commandline in the FileName property.

Instead, you should just Start the txt file, which will open in the user's default editor:

Process.Start(@"C:\Users\Karthick\AppData\Local\Temp\5aau1orm.txt");

As mentioned by @SLaks, this is the appropriate way let the default application (in your case mapped to the .txt extension) open the file

Process.Start("test.txt");

But if you prefer to open text files only in Notepad and not in other default Text Editor

ProcessStartInfo processStartInfo = new ProcessStartInfo(@"c:\Windows\System32\notepad.exe", "text.txt");
Process.Start(processStartInfo);

You are trying to execute c:\\windows\\system32\\notepad.exe C:\\Users\\Karthick\\AppData\\Local\\Temp\\5aau1orm.txt. If you aren't using a shell, it will be interpreted literally. If you use the shell, the shell will take care of the argument parsing.

Use the ProcessStartInfo.Arguements property to supply arguments instead.

FileName property dosen't take command line type syntax. What you have specified is for command line.

Since its only a .txt file you can use Process.Start() Method with full file path. It will automatically search for the respective default program to open the file.