arguments in awk command

Question

for var in 1 2 3 4 5 6 7 8 9 10
do
echo $var
export lower_bound=$((($var - 1) * 10))
echo $lower_bound
export upper_bound=$(($var * 10))
echo $upper_bound
awk '$2 >= $(($lower_bound)) && $2 < $(($upper_bound))' shubham_test.txt > file_$var.txt
#awk -v var1="$lower_bound" -v var2="$upper_bound" '{$2 >= $var1 && $2 < $var2}' shubham_test.txt > file_$var.txt
done

I am trying to split a file based on the value of a field in a text file in unix. The awk works when I pass hardcoded values to caompare but does not work properly if I am passing varaibles for comparison (upper_bound and lower_bound).

I looked it up and even replaced the awk command with the following:

awk -v var1="$lower_bound" -v var2="$upper_bound" '{$2 >= $var1 && $2 < $var2}' shubham_test.txt > file_$var.txt

so that it takes arguments. But this is not working either. Can anyone help?

Answer 1

Your second format should work, but without the $ for the variable names - so :

awk -v var1="$lower_bound" -v var2="$upper_bound" '{$2 >= var1 && $2 < var2}' shubham_test.txt

Your first version fails because the single quotes mean the unix shell does not evaluate the $lower_bound variables, but instead is passing the entire literal string to awk. If you're writing shell scripts, you must know how to use single-, double- and back-quotes - take a look at http://www.tutorialspoint.com/unix/unix-quoting-mechanisms.htm for example.