Bash check if array of variables have values or not

Question

I have an array of variables. I want to check if the variables have a value using the for loop.

I am getting the values into loop but the if condition is failing

function check {
    arr=("$@")
    for var in "${arr[@]}"; do
        if [ -z $var ] ; then
            echo $var "is not available"
        else
            echo $var "is available"
        fi
    done
}

name="abc"
city="xyz"
arr=(name city state country)
check ${arr[@]}

For the above I am getting all as available

Expected output is

name is available
city is available
state is not available
country is not available

`var` in your loop takes on the values _name_, _city_, _state_ and _country_. [ -z ... ] tests whether the length of the argument is zero. None of those words has zero length, so the `else` branch is taken every time. — user1934428, Feb 28 '20 at 11:42

Ivan · Accepted Answer · 2020-02-28T10:44:59.970

4

This is the correct syntax for your task

if [ -z "${!var}" ] ; then
    echo $var "is not available"
else
    echo $var "is available"
fi

Explanation, this method uses an indirect variable expansion, this construction ${!var} will expand as value of variable which name is in $var.

Changed check function a bit

check () {
    for var in "$@"; do
        [[ "${!var}" ]] && not= || not="not "
        echo "$var is ${not}available"
    done
}

And another variant using declare

check () {
    for var in "$@"; do
        declare -p $var &> /dev/null && not= || not="not "
        echo "$var is ${not}available"
    done
}

From declare help

$ declare --help
declare: declare [-aAfFgilnrtux] [-p] [name[=value] ...]
    Set variable values and attributes.
    Declare variables and give them attributes.  If no NAMEs are given,
    display the attributes and values of all variables.
    ...
    -p  display the attributes and value of each NAME
    ...

Actually all vars can be checked at once using this

check () {
    declare -p $@ 2>&1 | sed 's/.* \(.*\)=.*/\1 is available/;s/.*declare: \(.*\):.*/\1 is not available/'
}

edited Feb 28 '20 at 10:44

answered Feb 28 '20 at 09:42

Ivan

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It would be great if you can explain the change. – Fazlin Feb 28 '20 at 09:44
-z string True if the length of string is zero. – bob dylan Feb 28 '20 at 09:51
This has a bug - variables with values with spaces will result in an error message from `test`. – KamilCuk Feb 28 '20 at 09:51
If space isn't valid value than it's not a bug it's a feature) – Ivan Feb 28 '20 at 09:59
And i just followed OPs style – Ivan Feb 28 '20 at 10:01

score 0 · Answer 2 · answered Feb 28 '20 at 10:53

While indirection is a possible solution, it is not really recommended to use. A safer way would be to use an associative array:

function check {
    eval "declare -A arr="${1#*=}
    shift
    for var in "$@"; do
        if [ -z "${arr[$var]}" ] ; then
            echo $var "is not available"
        else
            echo $var "is available"
        fi
    done
}

declare -A list
list[name]="abc"
list[city]="xyz"

check "$(declare -p list)" name city state country

This returns:

name is available
city is available
state is not available
country is not available

The following question was used to create this answer: How to rename an associative array in Bash?

Bash check if array of variables have values or not

2 Answers2