CODEWITHSUNDEEP
bashfunctionshell
Droïdeka
Droïdeka

Reputation: 47

BASH pass argument to function doesn't work

I try to pass an argument to the function install() but the output is "add" and not my argument. I don't know how to get just the REP when i call the function directory-install(), because for know i have all the phrase "add directory...."

function install () {

    echo $1 #output not ok, show "add"

}


function directory-install () {

    read -p "enter directory " REP
    if cd $REP 2> /dev/null ; then
        echo -e "add directory '$REP'\n"
    else
        mkdir REP
        echo -e "creat directory \n"
    fi
    echo $REP
}


REP=$(directory-install)
echo $REP #output not ok too show "add directory..." but i just want the REP .
install $REP

Upvotes: 0

Views: 74

Answers (1)

Mr. Llama
Mr. Llama

Reputation: 20909

Your install function is working 100% correctly, the problem is that you're passing it garbage. Garbage in, garbage out.

The real culprit is your directory-install function. When you execute REP=$(directory-install), the variable REP now contains all of the text output from directory-install, not just the final echo $REP. That means the add directory or creat directory text.

If you want directory-install to only return REP, then you need to make sure that you have no other output in the function call. Alternately, you can redirect the non-return text to STDERR where it will be displayed, but not captured.

Example:

function badExample() {
    echo "Hello World"
    echo "ReturnText"
}

rtn=$(badExample)
# rtn now contains "Hello World\nReturnText"


function goodExample() {
    echo "Hello World" 1>&2
    echo "ReturnText"
}

rtn=$(goodExample)
# "Hello World" will appear on screen and rtn contains "ReturnText"

Upvotes: 1

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