How to test if an executable exists in the %PATH% from a windows batch file?

for %%X in (myExecutable.exe) do (set FOUND=%%~$PATH:X)
if defined FOUND ...

If you need this for different extensions, just iterate over PATHEXT:

set FOUND=
for %%e in (%PATHEXT%) do (
  for %%X in (myExecutable%%e) do (
    if not defined FOUND (
      set FOUND=%%~$PATH:X
    )
  )
)

Could be that where also exists already on legacy Windows versions, but I don't have access to one, so I cannot tell. On my machine the following also works:

where myExecutable

and returns with a non-zero exit code if it couldn't be found. In a batch you probably also want to redirect output to NUL, though.

Keep in mind

Parsing in batch (.bat) files and on the command line differs (because batch files have %0–%9), so you have to double the % there. On the command line this isn't necessary, so for variables are just %X.

Windows Vista and later versions ship with a program called where.exe that searches for programs in the path. It works like this:

D:\>where notepad
C:\Windows\System32\notepad.exe
C:\Windows\notepad.exe

D:\>where where
C:\Windows\System32\where.exe

For use in a batch file you can use the /q switch, which just sets ERRORLEVEL and doesn't produce any output.

where /q myapplication
IF ERRORLEVEL 1 (
    ECHO The application is missing. Ensure it is installed and placed in your PATH.
    EXIT /B
) ELSE (
    ECHO Application exists. Let's go!
)

Or a simple (but less readable) shorthand version that prints the message and exits your app:

where /q myapplication || ECHO Cound not find app. && EXIT /B