UNIX for Dummies Questions & Answers

I know I can obtain the file names in a folder by doing the following.

Code:

ls | awk '{t=$1; print t}'

Rather than print t, I'd like to take the file name as an input to a script like

Code:

bash-3.00$ more test.sh
echo "this is a test"
ls -l $1

When I run the script, I put "./test.sh a.out" and the script simply does "ls -l a.out".

I'm trying to combine the two things together and tried

Code:

ls | awk '{t=$1; test.sh t)}'

It doesn't work. What is the right way to do it? Thanks.

Actually what is your goal?

You can do this

Code:

for flnm in `ls -1`
do
./test.sh $flnm
done

My goal is to find out what files are in a folder and take each one of them as an input to a command. In my example, I simplified the command (used ls -l) for discussion purpose. Also I put the command in a script.

I can do an "ls" and manually process each file like

./test.sh file1
./test.sh file2
....
./test.sh file100

I'm trying to find a way to run this automatically. Just saw grep_me's response after I typed the above. Will try. Thanks.

Here is the general way to do that:

Code:

for file in *; do
  command $file
done

For many commands, you can simply say:

Code:

command *

Also if you want, put a test to verify if file is a regular file to skip directories or non-regular files:

Code:

for file in *
do
        [ -f "$file" ] && /path/test.sh "$file"
done