Question

I am trying to write a windows batch file that will print and execute a command if a window is open on the system. I can do this with a process but this is not the effect i want. I want to know if a window is open if so then print open if not do nothing, If there is a way that i can do this please let me know. Better description is when you open taskmanager and the first tab shows open windows. I just want a list of these windows and then from there just search for a program such as googlechrome.exe

Thank you all in advanced!

Answer 1

To see which executables are running from the command line, use: tasklist.exe. This enables you to filter on a WINDOWTITLE attribute as well. Beyond that, you might be out of luck.

Example to see if an untitled notepad instance is running:

tasklist /FI "WINDOWTITLE eq Untitled - Notepad"

You might use this inside a batch file as follows:

FOR /F "tokens=* USEBACKQ" %%F IN (`tasklist /FI "WINDOWTITLE eq Untitled - Notepad" ^| find/c "exe"`) DO (
SET windowcount=%%F
)

echo %windowcount%

if "%windowcount%"=="0" (
echo no windows!
) ELSE (
echo %windowcount% windows found!
)