Bash arrays: Need help with white space

Question

I'm trying to create an array in bash from a file with the following sample format:

data, data, interesting
data, data, more interesting

The way I'm populating the arrays:

read -r -a DATA1 <<< $(cat $FILE | awk -F, '{ print $1 }')
read -r -a DATA2 <<< $(cat $FILE | awk -F, '{ print $2 }')
read -r -a DATA3 <<< $(cat $FILE | awk -F, '{ print $3 }')

When I examine the array DATA3, there are 3 elements:

interesting
more
interesting

I need it to show only 2 elements like:

interesting
more interesting

How can I preserve the white space in field 3 so when I call that element from the array, it appears as "more interesting"? Is there a better way to handle this?

Answer 1

The key is to use the IFS (internal field separators) variable together with read.

read can read words directly into an array using the -a option. Set IFS=, and it will split at comma:

while read f; do
    echo "$f" | (IFS=, && read -a arr && echo ${arr[2]})
done

will echo

interesting
 more interesting 

You can also read directly into variable names:

IFS=, && read f1 f2 f2

EDIT: I can recommend reading Advanced Bash Scripting Guide.

Answer 2

Use cut. Example:

read -r -a DATA1 <<< $(cut -d, -f 1 $FILE)
read -r -a DATA2 <<< $(cut -d, -f 2 $FILE)
read -r -a DATA3 <<< $(cut -d, -f 3 $FILE)