How to pass variable name to a function and get it available outside in bash?

Question

How can I pass a variable name to a function and use it outside its scope?

I tried this but it doesn't work:

questionPrompt() {
    while [ -z "$answer" ]; do
    echo $1
    read answer
    ${2}=$answer
    done
}

questionPrompt "Which color do you like?" "COLOR"

echo $COLOR

It says: COLOR=red: command not found

Why would you want to create a function that does exactly the same thing as `read -p "question" answervar`? — Léa Gris, Jul 13 '20 at 17:10

score 1 · Accepted Answer · answered Jul 13 '20 at 16:50

1

Use declare:

declare -g "$2=$answer"

or

declare -gn var=$2
var="$answer"

However, the version of bash that ships with macOS (3.2) doesn't support declare -g; you can use printf instead.

printf -v "$2" '%s' "$answer"

answered Jul 13 '20 at 16:50

chepner

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It seems to work but it put me on exit and I can run anymore any other function call e.g. `questionPrompt "Which color do you like?" "SIZE"` – KaMZaTa Jul 13 '20 at 17:05
`answer` is global as well, so the second time you call it, `answer` is already non-empty, so the loop is never entered. Add `local answer` as the first line of the function. – chepner Jul 13 '20 at 17:09

How to pass variable name to a function and get it available outside in bash?

1 Answers1