How can I pass a variable into parameter expansion?

Question

I'm checking whether or not environment variables are set.

This gives me exactly what I want in the singular:

 : ${API_KEY:?`echo "Need to set API_KEY"`}

However, when I extract a function to loop over a collection of variables, things don't work like I'd expect.

 function check_env_for { 
   : ${$1:?`echo "Need to set $1 non-empty"`}
 }
 check_env_for API_KEY

breaks, returning:

 -bash: ${$1:?`echo "Need to set $1 non-empty"`}: bad substitution

While a solution is appreciated, what core shell fundamentals am I likely missing?

chepner · Accepted Answer · 2013-12-18 17:07:39Z

1

Along the lines of what you are trying, you can use indirect parameter expansion. This does require a temporary variable, which shouldn't be much of a hardship as this is a function definition:

function check_env_for {
    local foo="$1"
    : ${!foo:?"Need to set $foo non-empty"}
}

However, note that the error message isn't ideal:

$ check_env_for FOO
bash: !foo: Need to set FOO non-empty

Since you are writing a function, it's probably clearer to skip the "Display Error" parameter expansion and just test for a non-empty string, again using indirect parameter expansion:

check_env_for () {
    [[ -z ${!1} ]] && printf "Need to set $1 non-empty\n"
}

edited Dec 18, 2013 at 17:07

answered Dec 18, 2013 at 16:52

chepner

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2 Comments

Donovan Over a year ago

This produces an error locally. check_env_for:1: bad substitution

chepner Over a year ago

Are you sure you are using bash? Works for me in 3.2.51

Donovan · Accepted Answer · 2013-12-18 16:54:00Z

0

I got it working with the following code, which you can modify to suit your needs:

function check_env_for { 
   KEY="$(eval echo \$$1)"
   echo ${KEY:-"Need to set $1 non-empty"}
}
check_env_for API_KEY

Here it is in one line:

function check_env_for { 
  echo ${$(eval echo \$$1):-"Need to set $1 non-empty"}
}

This returns either the value of $API_KEY or Need to set API_KEY non-empty

edited Dec 18, 2013 at 16:54

answered Dec 18, 2013 at 16:35

Donovan

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5 Comments

Matthew Over a year ago

This doesn't echo anything, whether or not the parameter exists as a variable.

chepner Over a year ago

Don't use eval to re-implement indirect parameter expansion.

Donovan Over a year ago

The eval is necessary to obtain the value of the the ENV variable in this context. This is what eval is for.

chepner Over a year ago

If you are using bash, eval is not necessary. Using eval is never a good idea, since it will evaluate whatever is in the expanded variable, not just what you assume the contents will be.

Donovan Over a year ago

eval was created for just such purposes. Yes, I agree it's not always the best solution, particularly if you don't control your environment, but in this case, based upon the OP, I think it's the right solution. Sometimes simple is best.

alvits · Accepted Answer · 2013-12-18 18:01:50Z

0

Using ${parameter:?word} expansion does the trick but the output looks unpleasant. Using ${parameter:-word} expansion does look better.

If you still prefer the former, your code should look like this.

#!/bin/bash
function check_env_for {
        eval : \${$1:?"Need to set $1 non-empty"}
}
check_env_for API_KEY

Give this one a try and see if you like it.

#!/bin/bash
function check_env_for {
        eval echo \${$1:-"Need to set $1 non-empty"}
}
check_env_for API_KEY

The eval in both code will evaluate $1 and expand it to API_KEY. The $ outside of {} is escaped therefore it will not be expanded on eval.

answered Dec 18, 2013 at 18:01

alvits

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