How to substitute a string in bash script

Question

Note:

• Bash 3.00

How to substitute this example string 123456789, to look like 123-456-789

#!/bin/sh
# trivial example
read number;
# monotically substitute '-' into string after first three and dix digits

Answer 1

phone=`echo $phone | sed 's/\(...\)\(...\)/\1-\2-/'`

Answer 2

one way with gawk

$ echo "123456789" |awk  'BEGIN{FS=""}{ for(i=1;i<=NF;i+=3)s=s$(i)$(i+1)$(i+2)"-";sub(/-$/,"",s);print s } '
123-456-789

$ echo "123456789abcdef" | awk  'BEGIN{FS=""}{ for(i=1;i<=NF;i+=3)s=s$(i)$(i+1)$(i+2)"-";sub(/-$/,"",s);print s } '
123-456-789-abc-def

Answer 3

Without the use of sed:

$ number=123456789
$ number=${number:0:3}-${number:3:3}-${number:6:3}
$ echo $number
123-456-789