Creating Bash Array with Multiple Elements

Question

Very simple bash problem. I have an array that look like:

my_array=(1 2 3)

However, when I print the array or loop through it, bash only refers to the first element.

echo $my_array
1

for element in my_array ; do
    echo $element
done
1

How can I access all the elements?

If you add a Bash shebang, Shellcheck identifies the problems with the code snippets. See (the first paragraph of) the Stack Overflow 'bash' Info page for more about ShellCheck. — pjh
– pjh, Commented May 28, 2019 at 15:44

choroba · Accepted Answer · 2019-05-28 16:15:44Z

4

You need to use the proper syntax. To display all the elements, use

for element in "${my_array[@]}" ; do
    printf '%s\n' "$element"
done

$my_array is the same as ${my_array[0]}.

Also, the loop in the original question outputs my_array, not 1. Without a $, it's not a variable, it's just a word.

answered May 28, 2019 at 14:33

choroba

245k27 gold badges221 silver badges304 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1