getting filename part of a path in windows batch script

Question

In a batch script, I have a batch var %filePath% that has the path to a file. If I want to extract just the filename part from it, what is the command for that?

Thanks!

Answer 1

try this:

 for /f "delims=" %%i in ("%filePath%") do set "fileName=%%~nxi"
 echo %fileName%