How to pass the file name dynamically to awk when awk is used in command substitution to build the shell array?

Question

Here is the script I am using and this is working fine:

# This is a test file which we are creating and this will be used to create the shell array
cat > test.txt <<End-of-message
1A|1B|1C|1D
2A|2B|2C|2D
3A|3B|3C|3D
4A|4B|4C|4D
End-of-message

set -A col1_arr `awk 'BEGIN { FS = "|" }{print $1}' test.txt`

i=0
while [ $i -lt ${#col1_arr[@]} ]
do
    echo "Column 1 is:${col1_arr[$i]}"
    (( i=i+1 ))
done

rm test.txt # delet the test file

I do not want to hard-code test.txt and want to be able to pass it as a variable. Please advise.

Answer 1

set -A col1_arr `awk -v file=$myfile -F '|' \
    'END { while (getline < file) print $1 }' /dev/null`

Awk will advance to a new record only when the main input stream has a new record. So you have to do the loop manually in an END block, and make sure the awk script reaches the end block by specifying /dev/null as the main input stream.

Answer 2

It seems you are trying to produce the following output:

Column 1 is: 1A
Column 1 is: 2A
Column 1 is: 3A
Column 1 is: 4A

If that is the case, we don't need that long and complicated series of commands. This will do:

awk -F'|' '{print "Column 1 is: " $1}' test.txt

If you still want to use a variable instead of file name, then set it (example in bash):

fileName=test.txt
awk -F'|' '{print "Column 1 is: " $1}' $fileName