Using variables in loop bash

Question

I have four variables like:

a1="11"
a2="22"
b1="111"
b2="222"

and now I want to create a loop to check if it is null or not:

for (( i=1; i<=2; i++)); do
    eval "a=a$i"
    eval "b=b$i"
    if [ -z $a ] || [ -z $b ]; then
        echo "variable-$a: condition is true"
    fi
done

I want to count variable a and b then print the content of that. But in this way doesn't work and it checks:

a1
a2
b1
b2

But I need to check:

There is the number zero (0), there are empty variables (""), there are unset variables. But what is null supposed to be? — ndim
– ndim, Commented Jun 30, 2020 at 9:05
Could these be passed as one variable of the form "11 22 111 222" instead? — tadman
– tadman, Commented Jun 30, 2020 at 9:08

KamilCuk · Accepted Answer · 2020-06-30 09:17:10Z

1

Use variable indirection expansion:

for (( i=1; i<=2; i++)); do
    a=a$i
    b=b$i
    if [ -z "${!a}" ] || [ -z "${!b}" ]; then
        echo "variable-$a: condition is true"
    fi
done

But in this way doesn't work and it checks:

Because you never expanded the variables, there is no point in eval in your code. Your code is just:

for (( i=1; i<=2; i++)); do
    a="a$i"
    b="b$i"
    # Remember to quote variable expansions!
    if [ -z "$a" ] || [ -z "$b" ]; then
        echo "variable-$a: condition is true"
    fi
done

while you could:

for (( i=1; i<=2; i++)); do
    eval "a=\"\$a$i\""
    eval "b=\"\$b$i\""
    if [ -z "$a" ] || [ -z "$b" ]; then
        echo "variable-a$i: condition is true"
    fi
done

but there is no need for evil eval.

answered Jun 30, 2020 at 9:17

KamilCuk

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