Assign each line of file to be a variable

Question

I am looking to assign each line of a file, through stdin a specific variable that can be used to refer to that exact line, such as line1, line2

example:

cat Testfile
Sample 1 -line1
Sample 2 -line2
Sample 3 -line3

Answer 1

The wrong way to do this, but exactly what you asked for, using discrete variables:

while IFS= read -r line; do
    printf -v "line$(( ++i ))" '%s' "$line"
done <Testfile
echo "$line1" # to demonstrate use of array values
echo "$line2"

The right way, using an array, for bash 4.0 or newer:

mapfile -t array <Testfile
echo "${array[0]}" # to demonstrate use of array values
echo "${array[1]}"

The right way, using an array, for bash 3.x:

declare -a array
while read -r; do
  array+=( "$REPLY" )
done <Testfile

See BashFAQ #6 for more in-depth discussion.

Answer 2

bash has a builtin function to do that. readarray reads lines from a stdin (which can be your file) and assigns them elements of an array:

declare -a lines
readarray -t lines <Testfile

Thereafter, you can refer to the lines by number. The first line is "${lines[0]}" and the second is "${lines[1]}", etc.

readarray requires bash version 4 (released in 2009), or better and is available on many modern linux systems. Debian stable, for example, currently provides bash 4.2 while RHEL6 provides 4.1. Mac OSX, though, is still usingbash 3.x.