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.
