3

I have this small code snippet in a shellscript to set a password:

if [[ -z "${PASS+x}" ]]; then
    read -p "enter pass or use default [test1234]" PASS
    if [[ -z "$PASS" ]]; then
        PASS="test1234"
    fi
fi


if [ $EUID != 0 ]; then
    sudo "$0" "$@"
    exit $?
fi

When I run the script the read command is executed twice. This isn't supposed to happen. Can anybody help me with finding the error.

complete script: github (corrected already)

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7
  • read command shouldn't be executed twice, because you call it one time. And there's no loop it your code. Commented Feb 19, 2015 at 16:20
  • that's what I thought :/ . I tried changing it to echo "enter ...." read -s PASS, but its still the same. I have no idea where the error is. Commented Feb 19, 2015 at 16:23
  • @webKnjaZ I added a link to the complete script. But I think there shouldn't be anything relevant to the error? Commented Feb 19, 2015 at 16:28
  • Can you run the script with -x option and paste the output. If read is being invoked twice, it will show up. Commented Feb 19, 2015 at 16:55
  • @Ketan thanks for your help, I just found the solution godlygeek posted (didn't know about the -x switch - good to know for debugging!). Commented Feb 19, 2015 at 17:06

1 Answer 1

Reset to default
2

In your full script that you link to, you have this:

if [ $EUID != 0 ]; then 
    sudo "$0" "$@" 
    exit $? 
fi

That's done after all of the options handling and the read call that you show in your question. My guess is that you're hitting this code block, reinvoking the script using sudo, and hitting the read command twice - once before restarting the script using sudo, and once after.

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1
  • Thanks, I just found it at the moment. (while using the -x argument) Commented Feb 19, 2015 at 17:05

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