If I understood your situation correctly, I believe the solution is to use eval to pass the result of the command substitution though the shell's parser.
I'm using an old version of bash which does not support "set -A", so I'm using positional parameters instead of an array variable to try and explain myself. The principles involved are identical. Hopefully, I am able to get my point across
Code:
$cat eval.sh
# The goal is to assign the value of $first to $1
# and the value of $second to $2.
first='$second' second='boo far'
# 1)
# This first example exhibits the problem you are
# experiencing. The quotes intended to delimit the
# string are taken as part of the string itself,
# since command substitution occurs after the parsing
# step which cares about quoting.
set -- `echo \"$first\" \"$second\"`
echo "First positional parameter: $1"
echo "Second positional parameter: $2"
# Outputs:
# First positional parameter: "$second"
# Second positional parameter: "boo
# -----------------------------------------
# 2)
# Adding eval passes the result from command substitution
# through the parser, and the quotes are indeed interpreted
# as you'd like. However, with weak double quotes, if
# there is anything in the substituted text that would
# cause the shell to attempt parameter/command substitution,
# the output will be undesirably affected. In this case, the
# value of $first, the string "$second", is taken as a parameter
# which is then substituted, resulting in the value of $1
# being identical to that of $2.
eval set -- `echo \"$first\" \"$second\"`
echo "First positional parameter: $1"
echo "Second positional parameter: $2"
# Outputs:
# First positional parameter: boo far
# Second positional parameter: boo far
# -----------------------------------------
# 3)
# The correct solution is wrap the command substituted values
# in strong quotes, to ensure that they are taken literally and
# not word split when passed through the parser a second time.
eval set -- `echo \'$first\' \'$second\'`
echo "First positional parameter: $1"
echo "Second positional parameter: $2"
# Finally, the correct output:
# First positional parameter: $second
# Second positional parameter: boo far
Hope this helped,
alister
Last edited by alister; 01-04-2010 at 10:52 PM..
Reason: typos
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01-04-2010
