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Executing all scripts in /DIR except one

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# 1  
Old 12-25-2009
Executing all scripts in /DIR except one
First i need to find all scripts directly under /DIR that end with ".sh" extension except "noallow.sh". That can be done with:

Code:
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:

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  
Old 12-25-2009
It can even be done without find:

ksh (and zsh with kshglob enabled):

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

For bash you should enable extended glob with:

Code:
shopt -s extglob

Regarding your question for find/exec:

HTML Code:
find "$path" -maxdepth 1 -name '*.sh' ! -name 'noallow.sh' -exec {} \;
# 3  
Old 12-25-2009
Quote:
Originally Posted by proactiveaditya
Code:
for filename in $(find /DIR -maxdepth 1 -name "*.sh"|grep -v "noallow.sh")
do
/DIR/${filename}
done


Code:
dir=/path/to/wherever
for script in "$dir"/*.sh
do
  case $script in "$dir"/noallow.sh) continue ;; esac
  "$script"
done

# 4  
Old 12-25-2009
--others had posted the same--
# 5  
Old 12-25-2009
Code:
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  
Old 12-25-2009
Quote:
Originally Posted by proactiveaditya
Code:
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?

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

Last edited by cfajohnson; 12-25-2009 at 04:51 PM..
# 7  
Old 12-25-2009
Thanks a lot
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