C# Why does Process.Start("notepad.exe" myFile) is working and Process.Start("notepad++.exe" myFile) is not working

Question

The code, in both case is identical:

This is working and opening the text file in notepad

editor = "notepad.exe";
if (File.Exists(briefingFile))
{
  Process.Start(editor, briefingFile);
}

This one does is not work:

editor = "notepad++.exe";
if (File.Exists(briefingFile))
{
  Process.Start(editor, briefingFile);
}

It is the same test file and I have notepad++ installed. I Also tried to specify notepad++ with full path but the result is the same. Instead of opening notepad++ I get the attached error messages which tries to create new file or open missing files.

Answer 1

The notepad.exe file is part of Windows and lives in the Windows folder, which is part of the default search path (an environment variable). Notepad++.exe is not part of Windows, and so its home folder is not part of the default search path.

Therefore, to open a process using Notepad++ you must also know the full path to the program.

When trying the full path, make sure you escape the folder separator characters properly, and you must make sure to account for spaces in your path. In this case, the reason you see the C:\Program error is because you haven't yet accounted for the space in Program Files.

editor = @"""C:\Program Files\Notepad++\notepad++.exe""";
try 
{
  Process.Start(editor, briefingFile);
}
catch(Exception ex)
{
   // Do something here
}

Also note how I switched to an exception handler instead of File.Exists(). Disk I/O is one of those rare places where you should prefer handling the exception. File.Exists() is particularly bad for this, and should be avoided.

---

One other option here is if you have enough control for your target machines to know for sure Notepad++ is even installed, then you also have enough control register it as the default program for the files types your using, meaning you can skip selecting a program name at all:

Process.Start(briefingFile);

Answer 2

The solution is to use double quotes in the text file, not in the application. In my case: change:

Process.Start(editor, briefingFile);

to

Process.Start(editor, $@"""{briefingFile}""");

The "editor" is a full path to the editor.

Answer 3

There is nothing wrong with your code, although, as Joel stated, there are drawbacks to using File.Exists(). You should only need to make sure that the Notepad++ folder is in your User/System Environment PATH variables. I added the path for my Notepad++ folder on this PC, which is just C:\Program Files\Notepad++\, and ran the same code and it opens the file in Notepad++ just fine.

I ran the below code in an empty .NET 3.1 forms project and it can execute just fine. Do you still get your error in a new project?

        OpenFileDialog fileDialog= new OpenFileDialog();
        fileDialog.Filter = "All Files (*.*)|*.*";
        fileDialog.FilterIndex = 1;
        string editor = "";
        if (fialDialog.ShowDialog() == DialogResult.OK)
        {
            editor  = fileDialog.FileName;
        }

        //Added the escaped quotes to the front & back of this in case the path contains spaces.
        var filePath = @"""C:\SOMELOCALFILEPATH\test.txt""";
        if (File.Exists(filePath))
        {
            Process.Start(editor, filePath);
        }

I set a break point and my editor is in this format when Process.Start executes: "C:\\Program Files\\Notepad++\\notepad++.exe"

Answer 4

I ran into this issue and had also exactly same two errors when I tried to open a .txt file which was installed under C:\Program Files (x86)\MyProgram\SecondFolder\MyTxt.txt

The problem was when I copied the file path from windows explorer and pasted it into Visual Studio it deleted the blank space between Files and (x86).

happens when pasted directly into VS to edit it later. Can be avoided if you paste the path into quotation marks

You have to check if the file path is correct. Otherwise it won't work.