Splitting bash command line argument

Question

Is this the best way to split up a colon separated bash command line argument?

#!/bin/bash
hostlist=`echo $1| awk '{split($0,Ip,":")} END{for (var in Ip) print Ip[var];}'`
for host in $hostlist
 do
  ....
 done

Answer 1

Another way would be to use IFS, the shell's built-in method to split strings into fields.

OLDIFS=$IFS
IFS=':'
set -f
for host in $hostlist; do
  set +f
  echo "$host"
done
set +f
IFS=$OLDIFS

set -f turns off filename generation (globbing): without it, wildcards *?\[ would be expanded in each word.

Answer 2

I think the simplest solution is just to use bash builtins:

#!/bin/bash
hostlist=${1//:/ }  # this will replace all : with a space

for host in $hostlist ; do
    echo ${host}
done

Another way (still simpler than your awk solution) is to use cut though this probably depends on GNU cut.

#!/bin/bash
hostlist=$(echo $1 | cut -d : --output-delimiter=" " -f 1-) 

for host in $hostlist ; do
    echo ${host}
done