Shell Programming and Scripting

I would like to write a script to do such thing , if a directory have file that the extension is today (yymmdd), then do a specific action , I will run a cron job to run this script .

For example

Today is 2nd , if the directory have files which file name are abc.121102 and def.121102 , then echo the file name , so the output is abc , def

tomorrow is 3rd , if the directory have files which the file name are xxx.121103 and yyy.121103 , then echo the file name , so the output is xxx , yyy

can advise how to it ?

thanks.

What shell are you running? bash has the simplest way to do this.

Code:

TODAY=`date +"%y%m%d"`

if [ `ls -l *.${TODAY} 2> /dev/null | wc -l` -ne 0 ]
then
        for file in *.${TODAY}
        do
                echo ${file} | awk -F"." ' { print $1 } '
        done
else
        echo "No files found."
fi

thx reply ,
it is a very good script .
if the file name is not abc.121102 , it is abc.program.121102 , what can I do ?

other file name is
def.program.121102
xxx.program.121102
yyy.program.121102

thanks.

Replace

Code:

echo ${file} | awk -F"." ' { print $1 } '

with

Code:

echo ${file} | sed 's/\.'$TODAY'//g'

It will work for abc.121102 or abc.program.121102

Thank you for reply and I have one more requirements , if I want to echo the first delimitor ( def , xxx , yyy ) of the file name , that mean when run the script , it display def , xxx ,yyy , what can i do ?

Thanks.

is it possible to do that ?

thanks.