Use file name as an input to a script

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# 1  
Use file name as an input to a script

I know I can obtain the file names in a folder by doing the following.
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
bash-3.00$ more
echo "this is a test"
ls -l $1

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

I'm trying to combine the two things together and tried
ls | awk '{t=$1; t)}'

It doesn't work. What is the right way to do it? Thanks.
# 2  
Actually what is your goal? Smilie
# 3  
You can do this
for flnm in `ls -1`
./ $flnm

# 4  
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

./ file1
./ file2
./ 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.
# 5  
My goal is to find out what files are in
a folder and take each one of them as
an input to a command
Here is the general way to do that:
for file in *; do
  command $file

For many commands, you can simply say:
command *

# 6  
Also if you want, put a test to verify if file is a regular file to skip directories or non-regular files:
for file in *
        [ -f "$file" ] && /path/ "$file"


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