Solved: Batch file issue - Program dettection

Hello again,

I am having an issue with a .bat that I have written. All it does is tell me which version of a program is installed. But I am trying to remember how to write the FOR statement so that if the program is not installed it show a message and not the error stating the registry entry does not exist. It is a simple FOR statement but I am having issues writing the Else portion of it. All I get with the FOR statements I write with the else do not even work...

Please help.

Below is the Working For statement.

FOR /F "tokens=2* delims= " %%A IN ('REG Query "HKLM\SOFTWARE\Microsoft\Windows\CurrentVersion\Uninstall\SnagIt7" /v DisplayVersion') DO Set _Snaginf=%%B
echo . = SnagIt Version %_Snaginf% is installed.


Please help.

Thanks again.

 

Code:

Set _Snaginf=
FOR /F "tokens=2* delims= " %%A IN ('REG Query "HKLM\SOFTWARE\Microsoft\Windows\CurrentVersion\Uninstall\SnagIt7" /v DisplayVersion 2^>Nul') DO Set _Snaginf=%%B
If "%_Snaginf%"=="" (Echo.Snagit is not installed) Else echo.SnagIt Version %_Snaginf% is installed.

The 2^>Nul will keep the ERROR: The system was unable to find the specified registry key or value. message from being displayed.

 

Thank you That really helps. How can I show an echoed message if a program is not found?

 

If the Snagit registry key is not found it displays Snagit is not installed. That's the Echo statement on the left side of the Else clause. Isn't that what you wanted?

 

Yes I see it now. Thank you.