How to find out if a batch file was opened by double click, or by command line

Question

On a Windows 7, I have an executable, say immutableProg.exe, which I want to call 3 times with certain parameters. This is done by the batch file myBatch.bat.

Content of myBatch.bat:

immutableProg.exe -a
immutableProg.exe -b
immutableProg.exe -c

The executable immutableProg.exe does have a special --keep switch which stops the executable from returning until a user hits any key. Now I want to add the --keep switch if and only if my batch myBatch.bat got double clicked like:

immutableProg.exe -a
immutableProg.exe -b
immutableProg.exe -c --keep

It shall NOT be added if a user calls the batch from commandline.

The question: How can I find out (from inside my batch's view) if it was opened by a double click or from command line?

Changing the default behavior of the immutableProg.exe is unfortunatelly not an option, neither is to give the batch file an extra parameter from commandline.

Answer 1

%cmdcmdline% gives the exact command line used to start the current Cmd.exe.

When launched from a command console, this var is "%SystemRoot%\system32\cmd.exe".

When launched from explorer (double clicked) this var is cmd /c ""{full_path_to_the_bat_file}"

Answer 2

To actually use the info in haxtbh's answer, you can do the following. It is not fool proof, but it usually works fine. It would take an unusual scenario for it to give a false reading.

echo %cmdcmdline%|find /i """%~f0""">nul && echo doubleClick || echo console launch