UNIX for Dummies Questions & Answers

I am trying to test arguments to see if they are files in any directory.

I have :
[ $# -f 0 ]

but it's not working

Is your script supposed to be invoked with the files you want to process, and terminate if they don't exist? $# indicates how many arguments there were (presumably you want to abort if there were none) but to check whether the named file(s) exist, you give them as arguments to "test -f".

Code:

case $# in 0) echo no files, try again >&2 ;; esac
for f in "$@"; do
  if [ -f "$f" ] ; then
    ... handle file
  else
    echo "$0: file $f not found -- skipping" >&2
  fi
done

when I run my program and enter 2 arguments at the command line it is not checking to see if the 2 arguments are files or not.

I have tried

if [ -f filename ]
elif [ -f $1 -o $2 ]

I guess I don't know how to make it so it uses the 2 arguments as filenames and then searches for them.

That's the loop. If you always want exactly two arguments and they both need to exist, you can do that too. The -o doesn't "remember" what you did before so you need to tell it again to look for a file.

Code:

if [ -f "$1" -o -f "$2" ]

The loop I posted earlier works for any number of file names, and checks each of them in turn; but if two need to exist at the same time, it won't check for that. (Of course, then you need -a "and", not -o "or".)

I need to check following three things:

a) a file exists but is zero byte
b) a file exists with non-zero bytes
c) a file doesn't exist at all

What switches do we use? Is it -f, -s, -z. Please explain the difference in these!

Many Thanks
Shalua

You know enough to be better served by finding the documentation and reading that. It's more precise and more authoritative than any of us here, on a typical day. Read man sh and maybe man test if you have that.

I am looking for an answer to my question. Any pointers or answers will be appreciated.

Thanks