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Subject: /bin/sh script not behaving as expected
From: Rick Miller <vmiller@hostileadmin.com>
To: FreeBSD Questions <freebsd-questions@freebsd.org>
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Hi all,

I have shell code whose purpose is to determine the first disk in the
system where FreeBSD is to be installed.  The code is not behaving as
expected and I=E2=80=99m hoping that fresh pairs of eyes might help me iden=
tify the
problem.

Here is the script along with an explanation of the implementation and
description of the problem:

#! /bin/sh

disks=3D"da2 da1 da0";

for d in ${disks}; do
   if [ -z "${disk}" -o "${disk}" '>' "${d}" ]; then
      : ${disk:=3D${d}};
   fi
done


echo $disk;


# Given the input(s), $disks, the expected behavior of the above code is to
# set $disk to "da0" which ends up being the chosen disk to install FreeBSD=
.
#
# The for() loop iterates over $disks.  The if() statement tests the status
of
# $disk; If $disk is unset/null, $disk is set to $d.  If $disk is set, it
# compares the binary value of $disk to $d and should select the element
with
# the lower binary value.  In this particular case, it is expected to selec=
t
# da0 as the ultimate value for $disk, but in practice, it appears to set
# $disk to da2 instead.
#
# NOTE: This is on FreeBSD 10.  The code has been tested in bash as well,
which
# returns the same results.

TIA

--=20
Take care
Rick Miller