How to evaluate an arithmetic expression in an "if" statement?

This is my script:

#!/bin/bash
num=1
while [ $num -lt 100 ]
do
if [ $num % 3 -eq "0" ]
then
echo "fizz"
elif [ $num % 5 -eq "0" ]
then
echo "buzz"
elif [ $num % 3 -eq "0" ] && [ $num % 5 -eq "0" ]
then
echo "fizzbuzz"
else
echo $num
fi ((num++))
done

It keeps saying bash: [: too many arguments.

2 Answers

The correct syntax is:

if [ $(($num % 3)) -eq "0" ]

From man bash:

Arithmetic Expansion

Arithmetic expansion allows the evaluation of an arithmetic expression and the substitution of the result. The format for arithmetic expansion is: $((expression))

The expression $num % 3 must be evaluated before it can be compared using -eq.

To expand on Eric's answer, you actually can forgo the square brackets if you're simply testing equality with respect to 0. You can also omit the "$" as parameter dereferencing is optional. For the statement in his example,

if [ $(($num % 3)) -eq "0" ]

is equivalent to

if ! ((num % 3))

making sure to add in the "!" operator.