Bash - Sort a list of strings

You can use xargs twice along with a built in sort command to accomplish this.

$ my_list='a z t b e c'
$ my_list=$(echo $my_list | xargs -n1 | sort | xargs)
$ echo $my_list
a b c e t z

If you permit using the sort program (rather than program a sorting algorithm in bash) the answer could be like this:

my_list='a z t b e c'
echo "$my_list" | tr ' ' '\n' | sort | tr '\n' ' '

The result: a b c e t z'

Arrays are more suitable to store a list of things:

list=(a z t b "item with spaces" c)

sorted=()
while IFS= read -rd '' item; do
    sorted+=("$item")
done < <(printf '%s\0' "${list[@]}" | sort -z)

With bash 4.4 you can utilize readarray -d:

list=(a z t b "item with spaces" c)

readarray -td '' sorted < <(printf '%s\0' "${list[@]}" | sort -z)

To use the array to create a simple menu with select:

select item in "${sorted[@]}"; do
    # do something
done

Using GNU awk and controling array traversal order with PROCINFO["sorted_in"]:

$ echo -n $my_list |
  awk 'BEGIN {
      RS=ORS=" "                            # space as record seaparator
      PROCINFO["sorted_in"]="@val_str_asc"  # array value used as order
  }
  {
      a[NR]=$0                              # hash on NR to a
  }
  END {
      for(i in a)                           # in given order
          print a[i]                        # output values in a
          print "\n"                        # happy ending
  }'
a b c e t z

You can do this

my_list=($(sort < <(echo 'a z t b e c'|tr ' ' '\n') | tr '\n' ' ' | sed 's/ $//\
'))

This will create my_list which is an array.