shell: write integer division result to a variable and print floating number - shell

Question

I'm trying to write a shell script and plan to calculate a simple division using two variables inside the script. I couldn't get it to work. It's some kind of syntax error.
Here is part of my code, named test.sh
awk '{a+=$5} END {print a}' $variable1 > casenum
awk '{a+=$5} END {print a}' $variable2 > controlnum
score=$(echo "scale=4; $casenum/$controlnum" | bc)
printf "%s\t%s\t%.4f\n", $variable3 $variable4 $score
It's just the $score that doesn't work.
I tried to use either
sh test.sh
or
bash test.sh
but neither worked. The error message is:
(standard_in) 1: syntax error
Does anyone know how to make it work? Thanks so much!

Answer 1

You are outputting to files, not to vars. For this, you need var=$(command). Hence, this should make it:
casenum=$(awk '{a+=$5} END {print a}' $variable1)
controlnum=$(awk '{a+=$5} END {print a}' $variable2)
score=$(echo "scale=4; $casenum/$controlnum" | bc)
printf "%s\t%s\t%.4f\n", $variable3 $variable4 $score
Note $variable1 and $variable2 should be file names. Otherwise, indicate it.

Answer 2

First your $variable1 and $variable2 must expand to a name of an existing file; but that's not a syntax error, it's just a fact that makes your code wrong, unless you mean really to cope with files containing numbers and accumulating the sum of the fifth field into a file. Since casenum and controlnum are not assigned (in fact you write the awk result to a file, not into a variable), your score computation expands to
score=$(echo "scale=4; /" | bc)
which is wrong (Syntax error comes from this).
Then, the same problem with $variable3 and $variable4. Are they holding a value? Have you assigned them with something like
variable=...
? Otherwise they will expand as "". Fixing these (including assigning casenum and controlnum), will fix everything, since basically the only syntax error is when bc tries to interpret the command / without operands. (And the comma after the printf is not needed).
The way you assign the output of execution of a command to a variable is
var=$(command)
or
var=`command`

Answer 3

If I understand your commands properly, you could combine calculation of score with a single awk statement as follows
score=$(awk 'NR==FNR {a+=$5; next} {b+=$5} END {printf "%.4f", a/b}' $variable1 $variable2)
This is with assumption that $variable1 and $variable2 are valid file names
Refer to #fedorqui's solution if you want to stick to your approach of 2 awk and 1 bc.