Error on retrieving array length in bash

Tag: bash,shell,sh

I've been trying to run a simple script that retrieves the penultimate subdirectory of a variable size directory. But it fails miserably to retrieve its length. The script goes:

ARRAY=$(ls files-0/var/log/)

echo ${#ARRAY[@]}

for i in ${#ARRAY[@]};
do
    echo $i
done

The output I get is:

If i change the line

for i in ${#ARRAY[@]};

for i in $ARRAY;

at least I get the list of directories/files:

I know there are lot's of posts approaching this question, but mostly all of them start from the basis of getting array length properly.

Best How To :

Following up on the comment, the syntax for initializing indexed array values in BASH is:

array=( element1 element2 etc... )

In your case you were simply assigning the result of the command substitution to the array with:

ARRAY=$(ls files-0/var/log/)

(which assigned the return of ls files-0/var/log/ to a variable named ARRAY)

The proper initialization of the ARRAY using command substitution is:

ARRAY=( $(ls files-0/var/log/) )

Note: when an array is declared with declare -a somename, or after it has been initialized with values, you can add an element to the array in the next open index with:

ARRAY+=( somevalue )  # quoted as necessary

As of BASH 4, you also have Associative Arrays (declare -A) which allow arrays to be created with key value pairs, instead of simple index value pairs. Where initialization is:

AARRAY=( [key1]=value1 [key2]=value2 etc... )

Good luck with BASH, it is a very capable shell.

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