Bash: If statement does not see the filename grab by loop

Question

I am trying to learn bash and I did a one liner like such:

for f in *; do echo $f; done

This prints the all the file name in current directory.

Output:

a file
testfile

Since f is able to grab "a file" just fine, then I would do something like this:

for f in *; do
    if [ -f ${f} ]; then
       LINE=$(wc -l < $f)
       WORD=$(wc -c < $f)
       echo $f ${LINE} ${WORD}
    fi
done

This script failed at line 4 because "a file" has space in them.

What I need help with: Why does it fail?

F is able to grab the filename properley. Intuitively, i would believe that -f ${f} would be able to check if "a file" is a file.

Why does that "a file" passed the for loop but failed at the if statement check?

Expanding the commandline is not done recursive, but only once. When `*` is expanded in filenames, these names are not split again on spaces. When you use `$f` in the loop the commandline is expanded again. Use quotes in the loop. — Walter A, Oct 11 '19 at 22:13
Shell syntax has a lot of sharp edges like this; I recommend [shellcheck.net](https://www.shellcheck.net) for pointing out common problems/mistakes. — Gordon Davisson, Oct 11 '19 at 22:16

markp-fuso · Accepted Answer · 2019-10-11T22:21:22.160

2

You need to (double) quote each reference of $f so that bash knows the 'a' and 'file' are not separate objects, eg:

for f in *; do
    if [ -f "${f}" ]; then
       LINE=$(wc -l < "${f}")
       WORD=$(wc -c < "${f}")
       echo "${f}" ${LINE} ${WORD}
    fi
done

edited Oct 11 '19 at 22:21

answered Oct 11 '19 at 22:11

markp-fuso

1

You should also double-quote the argument(s) to the final `echo` command. – Gordon Davisson Oct 11 '19 at 22:16
thanks, answer updated – markp-fuso Oct 11 '19 at 22:21

1 Answers1