how to pass the filename as variable to a awk command from a shell script - linux

Question

in my shell script i have the following line
PO_list=$(awk -v col="$1" -F";" '!seen[$col]++ {print $col}' test.csv)
which generates a list with the values from column "col" which came from "$1" from file test.csv.
it might be possible to have several files in same location and for this would need to loop among them with a for sentence. For this I have to replace the filename test.csv with a variable, $i for example, which is the index from the list of files.
trying to fulfill my request, I was modifying my line with
PO_list=$(awk -v col="$1" -F";" '!seen[$col]++ {print $col}' $j)
unfortunately, i receive the error message:
awk: cannot open test.csv (No such file or directory)
Can anyone tell me why this error occur and how can I solve it, please?
Thank you,

Answer 1

As you commented in your previous question, you are calling it with
abc$ ./test.sh 2
So you just need to add another parameter when you call it:
abc$ ./test.sh 2 "test.csv"
and the script can be like this:
PO_list=$(awk -v col="$1" -F";" '!seen[$col]++ {print $col}' "$2")
# ^^^^
Whenever you want to use other parameters, remember they are positional. Hence, the first one is $1, second is $2 and so on.
In case the file happens to be in another directory, you can replace ./test.sh 2 "test.csv" by something like ./test.sh 2 "/full/path/of/test.csv" or whatever relative path you may need.