How to run the "other" executable matching the same name found in PATH?

Question

Is there a concise way to call the "other" executable for find available in path that can be found with where?

C:\repos>where find
C:\Windows\System32\find.exe
C:\Program Files\Git\usr\bin\find.exe
C:\Users\Qwerty\cmd\UnxUtils_wbin\find.exe

For example something like find~2 or wannabe whereget find -n 2 -- <arguments for find here>

Answer 1

Calling an executable without the full path will always find and use the first matching filename by searching the directories in $PATH. For your example, if you always want to use the git find, simply ensure that the path to it appears in your $PATH before the Windows find. If this is unsatisfactory or impossible (e.g., due to policy), you should either create an alias in bash for it, or always call it using the full pathname.

Answer 2

I will just put it here

$ alias whereget='_whereget() { A=$1; B=$2; shift 2; eval \"$(where $B | head -$A | tail -1)\" $@; }; _whereget'
$ whereget 2 find . -type d

Unfortunately this solution only works in bash, not in Windows cmd.