Reading a variable value from an array of your variable names

Question

I wanted to try this approach in bash and it seems there is no straightforward way to do so. Example of what I was messing with:

Var1=foo
Var2=bar
Var3=

Var3 is deliberately left undefined in my tests with this.

After those variables, I use something ideally like this:

declare -a array=(Var1 Var2 Var3)

for array_element in ${array[@]};
do 
    if [[ $array_element ]];
        then 
            echo "Element $array_element has content - true";
            echo "The value of $array_element is foo."
            echo
        else 
            echo "Element $array_element has no content - false";
            echo
    fi
done

The desired end result I'm after to see if I can do it would be that if you had 3 or 30 possible variables, it would tell you if an array element has content or not, and then output that specific variable content. In this case, "foo" and "bar".

Define a variable
Define an array that includes the name of the variable only
Check to see if the defined array has content (not if it's set--just is it blank or not)
Then output the actual originally defined value of the array element.

Dynamic arrays, sort of? A basic bunch of

if [[ $Var1 ]] 
if [[ $Var2 ]] 
if [[ $Var3 ]]

works fine, but I want to wrap it up into something like this if I can to have less repetition of code. The long goal would be to have the variable be read, spit out whatever is defined as variables in a separate config file, and then fire off various functions defined elsewhere against that dynamic content. So, if this can be made to work, where I have up above:

echo "The value of $array_element is foo."

Would be replaced with a bunch of various functions() that would use foo as $1, essentially. But this part got me hitched up. My test above (unsurprisingly) fails as expected with all True hits, because it's just affirming $array_element is indeed set from the reading of the $array. I'm stuck on how to get the content that's defined up top under the $array_element. If you toss an "echo $Var1" anywhere in the file, it also outputs the expected "foo" regardless of where it is.

Can you do this in bash? I don't think I've seen variables used this way before and I'm not sure if I'm stuck on some mechanical problem/bash limitation or a logic problem in looking at it. I've been mucking around with it and a couple of approaches for a while with no luck, and Google and Stack Overflow searches have come up dry. Thanks for any help.

possible duplicate of Bash array expansion using variable indirection expansion — Barmar
– Barmar, Commented Aug 1, 2013 at 19:01

anubhava · Accepted Answer · 2013-08-01 19:12:06Z

0

This is the correct way of looping through an array that may contain empty elements:

Var1=foo
Var2=bar
Var3=
declare -a array=("$Var1" "$Var2" "$Var3")

for ((i=0; i<${#array[@]}; i++))
do 
    array_element="${array[$i]}"

    echo "processing [$array_element]"

    if [[ $array_element ]];
        then 
            echo "Element $array_element has content - true";
            echo "The value of $array_element is foo."
            echo
        else 
            echo "Element $array_element has no content - false";
            echo
    fi
done

OUTPUT:

processing [foo]
Element foo has content - true
The value of foo is foo.

processing [bar]
Element bar has content - true
The value of bar is foo.

processing []
Element  has no content - false

answered Aug 1, 2013 at 19:12

anubhava

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4 Comments

Lsrei Over a year ago

I was close, thank you -- but how about getting the output to show the specific initial variable? The value of Var1 is foo.

anubhava Over a year ago

Note this line: echo "processing [$array_element]" which is printing all values of your array. Pls note that you can only get specific element in array using a numeric index like 0, 1, 2 etc.

Armali Over a year ago

Replace $array_element by Var$((i+1)) in the echo command.

anubhava Over a year ago

@Lsrei: If this answer helped you solve your problem, please consider marking it as "accepted", so users facing a similar problem in the future will be able to see it easily.

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