splitting folder names with awk in a directory - linux

Question

There are some directories in the working directory with this template
cas-2-32
sat-4-64
...
I want to loop over the directory names and grab the second and third part of folder names. I have wrote this script. The body shows what I want to do. But the awk command seems to be wrong
#!/bin/bash
for file in `ls`; do
if [ -d $file ]; then
arg2=`awk -F "-" '{print $2}' $file`
echo $arg2
arg3=`awk -F "-" '{print $3}' $file`
echo $arg3
fi
done
but it says
awk: cmd. line:1: fatal: cannot open file `cas-2-32' for reading (Invalid argument)

Answer 1

awk expects a filename as input. Since you have said the cas-2-32 etc are directories, awk fails for the same reason.
Feed the directory names to awk using echo:
#!/bin/bash
for file in `ls`; do
if [ -d $file ]; then
arg2=$(echo $file | awk -F "-" '{print $2}')
echo $arg2
arg3=$(echo $file | awk -F "-" '{print $3}')
echo $arg3
fi
done

Answer 2

Simple comand: ls | awk '{ FS="-"; print $2" "$3 }'
If you want the values in each line just add "\n" instead of a space in awk's print.

Answer 3

When executed like this
awk -F "-" '{print $2}' $file
awk treats $file's value as the file to be parsed, instead of parsing $file's value itself.
The minimal fix is to use a here-string which can feed the value of a variable into stdin of a command:
awk -F "-" '{print $2}' <<< $file
By the way, you don't need ls if you merely want a list of files in current directory, use * instead, i.e.
for file in *; do

Answer 4

One way:
#!/bin/bash
for file in *; do
if [ -d $file ]; then
tmp="${file#*-}"
arg2="${tmp%-*}"
arg3="${tmp#*-}"
echo "$arg2"
echo "$arg3"
fi
done
The other:
#!/bin/bash
IFS="-"
for file in *; do
if [ -d $file ]; then
set -- $file
arg2="$2"
arg3="$3"
echo "$arg2"
echo "$arg3"
fi
done