command file: Opening a window only if not already opened

I am on a Windows 10 Home PC, ver. 21H1. I am a newbie to command files. I've written one to execute a program, and then display my local Temp folder. Unfortunately, if the Temp folder already is displaying in Explorer, the command opens a new instance of that window. How would I modify it to open the Temp window only if it has not already been opened, and otherwise, just give the open Temp window the focus? Thank you.

Here's my code:

@echo off
"C:\Program Files\progManufacturer\progName\prog.exe" -l
%SystemRoot%\explorer.exe %temp%
exit

I don't if this will work for you as I don't know the behaviour of your prog.exe ( seems to work for notepad when I tried it as replacement for prog.exe )

Batch file to start prog.exe, and then Explorer at %temp%, but only if it is not already open at %temp%

Code:

rem @echo off

start "C:\Program Files\progManufacturer\progName\prog.exe" -l
start %temp%
timeout 2
Powershell.exe -NoProfile -ExecutionPolicy Bypass -WindowStyle Minimized -command "Invoke-Item $env:TEMP"

pause

(If there are no errors, you can remove the rem in front of the echo command and the pause at the end. - if it doesn't work, hopefully one of the Gurus on Tenforums can delve deeper into it).

Thank you so much! It worked (although, frankly, I don't understand the powershell command). Here's my completed code:

"C:\Program Files\myProgManufacturer\MyProg\MyProg.exe" -l
start %temp%
timeout 1
Powershell.exe -NoProfile -ExecutionPolicy Bypass -WindowStyle Minimized -command "Invoke-Item $env:TEMP"
exit