Bash reference variable by string name

Question

I'm trying to write a simple function that checks whether a variables is declared and non empty or not.

my_var="dummy_val"

function validate_var(){
        ${1:?"Variable is not set"}
}

validate_var my_var
validate_var my_var2

I get the following error:

script: line n: my_var: command not found

Is there a simple way I could reference the variable from the function argument?

stackoverflow.com/questions/3601515/… -> stackoverflow.com/a/35412000/9072753 — KamilCuk
– KamilCuk, Commented Dec 21, 2021 at 12:55
Thanks, but I would like to solve the current issue, just for educational purposes. — Bogdan Prădatu
– Bogdan Prădatu, Commented Dec 21, 2021 at 13:05

choroba · Accepted Answer · 2021-12-21 13:14:52Z

1

This seems to work. It uses the ! variable indirection.

function validate_var(){
    : ${!1:?"$1 is not set"}
}

answered Dec 21, 2021 at 13:14

choroba

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2 Comments

Philippe Over a year ago

If var is set to empty string, it shows bash: !1: var is not set

choroba Over a year ago

I don't know what the expected output is for that case. Maybe validate the input first.

polo-language · Accepted Answer · 2021-12-21 12:50:56Z

1

Pass the variable by prepending a $. Also I've added an echo (as an example), otherwise your validate_var function tries to execute it's argument:

my_var="dummy_val"

function validate_var(){
        echo ${1:?"Variable is not set"}
}

validate_var $my_var

answered Dec 21, 2021 at 12:50

polo-language

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5 Comments

Bogdan Prădatu Over a year ago

With this solution I get "invalid variable name", even if the variable is declared as empty string (e.g. my_var2="").

Bogdan Prădatu Over a year ago

Also, this passes "dummy_val" to the function, instead of "my_var".

polo-language Over a year ago

It seems to work fine with an empty string: $ my_var2="" $ validate_var $my_var2 bash: 1: Variable is not set

polo-language Over a year ago

You mean you want to pass a variable by reference?

Bogdan Prădatu Over a year ago

Actually, you're right. It does work, I don't know what I was doing wrong. Anyway, @choroba's answer does exactly what I want. Thanks.

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