Q:

how to find and execute a file in bash script? 

Hi I am trying to us the find command and -exec to -execute a file file when found but it doesn't seem to work.  here is some code:
 ssh $i find $DIRECTORY -name "version.sh" -exec sh {} \;

I am getting the error no arguments for -exec
thanks! 

A:

Your side gets a semicolon, but the other side doesn't. 
... \\\;

