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  1. #1

    Bash - Variable with spaces and quotes passed to function - help!

    I have a generic script that builds up the parameters to call rsync. I am trying to add an optional set of parameters and pass this into a function that generates and executes the rsync command. The extra parameters I want to pass is --rsh="ssh -p 2000".

    So in my caller I have something along the lines of
    EXTRAPARAM="--rsh=\'ssh -p 2000'"

    Then I call the function
    testFunc $PARAM

    Running this shows (below) and seems to drop the arguments when it see the space.
    ++ ARG='-n --rsh=\'\''ssh'

    If I call testFunc "$PARAM" to keep the whole variable contents I get
    ++ ARG='-n --rsh=\'\''ssh -p 2020\'\'''

    As a test (not the real scripts) I wrote up the following to try various scenarios. Any one point me to the solution (if any?!)

    set -x

    testFunc() {

    local ARG1=$1
    ARG="$ARG $ARG1"

    rsync $ARG
    PARAM="--rsh=\'ssh -p 2000\'"

    testFunc $PARAM
    testFunc "$PARAM"

  2. #2
    How about a combination of escaping the double-quotes, and evaling it?, e.g.:
    testFunc() {
      opts="--rsh=\"ssh -p 22\""
    eval rsync $opts -av foo.txt remote_host:/tmp/

  3. #3
    Think that works actually!
    Last edited by ak888; 08-22-2011 at 03:42 PM. Reason: Thought it didn't work at first...

  4. $spacer_open

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