Executing all scripts in /DIR except one

Shell Programming and Scripting
1

First i need to find all scripts directly under /DIR that end with ".sh" extension except "noallow.sh". That can be done with:

find /DIR -maxdepth 1 -name "*.sh"|grep -v "noallow.sh"

Now i want to run all the files output from the previous command.

The following code:

for filename in $(find /DIR -maxdepth 1 -name "*.sh"|grep -v "noallow.sh")
do
/DIR/${filename}
done

Can do the job. But i want to know can this whole thing be done by using find only with exec {}.

2

It can even be done without find:

ksh (and zsh with kshglob enabled):

for s in "$path"/!(noallow).sh; do "$s"; done

For bash you should enable extended glob with:

shopt -s extglob

Regarding your question for find/exec:

find "$path" -maxdepth 1 -name '*.sh' ! -name 'noallow.sh' -exec {} \;
3 
dir=/path/to/wherever
for script in "$dir"/*.sh
do
  case $script in "$dir"/noallow.sh) continue ;; esac
  "$script"
done
4

--others had posted the same--

5 
find "$path" -maxdepth 1 -name '*.sh' ! -name 'noallow.sh' -exec {} \;

Now can i echo/print the name of the script before running it and all this using find command?

6 
find "$path"/*.sh ! -name noallow.sh -print -exec {} \;
7

Thanks a lot