How do you replace sed and wc with awk? (counting names in directories)

Question

I'd like to count the number of elements inside a folder. I thought of

function lsc { /bin/ls -l $1 | sed 1d | wc -l }

but then I remembered how awk people reduced those kind of pipes, getting rid of superfluous greps and seds. So, how would you do that in awk?

Answer 1

There is no need for ls, sed, wc or awk.

If you simply want to count how many names a pattern expands to, then you can do that with

set -- *
echo "$#"

The set command sets the positional parameters ($1, $2, etc.) to the names matching the * pattern. This automatically sets the special variable $# to the number of set positional parameters, i.e. the number of names matching the given pattern.

In bash or in a shell that has named arrays, you can use

names=(*)
echo "${#names[@]}"

This works similarly, but sets the elements of the names array to the names resulting from the expansion of the * pattern. The variable expansion ${#names[@]} will be the number of elements in the names array.

An issue with this is that if the pattern doesn't match anything, it will remain unexpanded, so you get a count of 1 (even though the directory is empty). To fix this in the bash shell, set the nullglob shell option with shopt -s nullglob. By setting this shell option, patterns that do not match anything will be removed completely.

In bash, if you additionally want to count hidden names, set the dotglob shell option with shopt -s dotglob.

Your function could look something like this in bash:

lsc () (
    shopt -s nullglob
    set -- "$1"/*
    echo "$#"
)

Note the use of ( ... ) for the function body to avoid setting nullglob in the calling shell.

Or, for /bin/sh:

lsc () {
    set -- "$1"/*
    if [ -e "$1" ] || [ -L "$1" ]; then
        echo "$#"
    else
        echo 0
    fi
}

The if statement here makes sure that the first positional parameter is the name of an actual file and not an unexpanded pattern (due to not matching anything). The -e ("exists") must be true for us to trust the number in $#. If it isn't true, then we additionally check whether the name refers to a symbolic link with the -L test. If this is true, we know that the first thing that the pattern expanded to was a "dead" symbolic link (a symbolic link pointing to a non-existent file), and we trust $# to be correct. If both tests fail, we know that we didn't match anything and therefore output 0.

Answer 2

First, your one-liner function definition is incorrect. Presence of the function keyword indicates that you might use Bash and this is what it says under Shell Function Definitions in man bash:

function name [()] compound-command [redirection]

while the compound command is defined as:

{ list; }

So it should be:

function lsc { /bin/ls -l $1 | sed 1d | wc -l; }

But even though it's technically correct it will not work well, see Why not parse ls (and what to do instead)?.

You can print a number of files and directories in the given directory using awk like that:

awk 'END {print ARGC - 1}' *

but notice that if at least one of the arguments expanded by shell in place of * is a directory awk complains:

$ awk 'END {print ARGC - 1}' *
awk: warning: command line argument `dir1' is a directory: skipped

But anyway, the result is still fine and you can redirect errors to /dev/null:

awk 'END {print ARGC - 1}' * 2>/dev/null