how to print environment variable value in shell of a variable whose name is present in another variable value

Question

I have store environment variable names in an array with name array and I want to check if it is set then only certain code should run. Like if somebody has set environment variable db_instance_type="t2.micro" then desire output should be -var="db_instance_type=$db_instance_type"

I have written below code but it is not working as expected because I am checking variable presence using single $ which will always be there and not checking main variable value.

set -u
db_instance_type="t2.micro" # for testing only
db_dns_ttl="300"    # for testing only
function check_input_vars() {
    local temp_cmd=''

    declare -a array=(db_instance_type db_vpc_id db_subnet_ids db_sg_cidr_allowed_range db_ebs_volume_size delete_on_termination_db_ebs_volume
                    security_group_id contact db_image_owner_id db_dns_ttl db_image_name_filter ws_key_name ws_dns_ttl ws_instance_type
                    ws_vpc_id  ws_subnet_ids ws_ami_id enable_ws_public_ip
                    )

    temp_cmd=''
    for var in "${array[@]}"
    do
        var_name=(${!var@})
        if [[ "${var_name}" ]]; then 
            temp_cmd+=" -var=\"$var=\$$var\""
        fi
    done   
    echo $temp_cmd
}

input_var_str="$(check_input_vars)"
echo "$input_var_str"  # output should be: -var="db_instance_type=$db_instance_type" -var="db_dns_ttl=$db_dns_ttl
unset db_instance_type  # for testing only

Fix:

I have to comment this line --> #set -u And I have used below code:

var_value=${!var}
if [[ ! -z "${var_value}" ]]; then temp_cmd+=" -var=\"$var=\$$var\""; fi

Subjectively: This sounds like a bad idea. I believe using so many global variables will only result in unmaintanable spaghetti code. unset db_instance_type # for testing only spagetti. — KamilCuk
– KamilCuk, Commented Sep 15, 2021 at 11:29
You want the output to be exactly literally -var="db_instance_type=$db_instance_type" or do you want the output ot be -var="db_instance_type=t2.micro"? You specified the first, so the command works like you want, so? Also why do you want to have " in the string - are you sure you want quotes preserved literally? — KamilCuk
– KamilCuk, Commented Sep 15, 2021 at 11:31
output should be: -var="db_instance_type=$db_instance_type" -var="db_dns_ttl=$db_dns_ttl — Ashish Sharma
– Ashish Sharma, Commented Sep 15, 2021 at 11:54

KamilCuk · Accepted Answer · 2021-09-15 12:14:35Z

1

The syntax is off - to get the variable value use ${!var}. No @. And ( ) are for array assignment - it's not an array, it's a single vlaue.

    var_value=${!var}
    if [[ -n "${var_value}" ]]; then..

Also, it accesses the value of the variable, where variable var stores the name of the variable, thus I would call the variable var_value.

Check your scripts with https://shellcheck.net .

To protect against set -u, expand variable to empty when not defined or empty.

 var_value=${!var:-}

Also: How to check if a variable is set in Bash?

output should be: -var="db_instance_type=$db_instance_type" -var="db_dns_ttl=$db_dns_ttl"

Related reading: https://mywiki.wooledge.org/BashFAQ/048 https://mywiki.wooledge.org/BashFAQ/050 https://mywiki.wooledge.org/BashFAQ/006 .

edited Sep 15, 2021 at 12:14

answered Sep 15, 2021 at 12:01

KamilCuk

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2 Comments

Ashish Sharma Over a year ago

I am not getting right output and also getting an error. !var: unbound variable (blank)

Ashish Sharma Over a year ago

Hi @kmilChuk, I figure out whats the issue here. The issue is` set -u` . When I removed it it work fine otherwise it fail.

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