What is the error in the if statement in this bash script?

Question

What is the problem with the following shell script? I'm getting

[: too many arguments

error

#!/bash
var1=10;
var2=20;
if [ $var1 % 2 -eq 0 ] -a [ $var2 %5 -eq 0 ];
then
    #something
fi

Answer 1

First, the [ command only does comparisons. You have to perform calculations separately. For example:

var1mod=$((var1 % 2))
if [ $var1mod -eq 0 ]; then

However, since you appear to be writing your script for bash, you can use a different command and avoid the extra step:

if (( var1 % 2 == 0 )); then

And second, you cannot chain commands with -a. This only works as an argument to [. But outside it, you must use &&.

Three different variants:

if [ $var1mod -eq 0 -a $var2mod -eq 0 ]; then
if [ $var1mod -eq 0 ] && [ $var2mod -eq 0 ]; then
if (( var1 % 2 == 0 )) && (( var2 % 2 == 0 )); then
if (( var1 % 2 == 0 && var2 % 2 == 0 )); then

Always remember, the syntax of if is a simple if command; then command; fi. This means that [ in if [ is just the name of a command like ls or echo.

Answer 2

You seem to be confusing Conditional Evaluation with Arithmetic Evaluation. % is an operator in Arithmetic Evaluation, not Conditional Evaluation. See the bash(1) man page sections for those topics. [...] is for Conditional Evaluation, whereas ((...)) is for Arithmetic Evaluation.

Not to mention that your shebang line is wrong, unless for some reason you installed a copy of bash at the root level of your boot drive.

Last but not least, a newline is the same as a semicolon. You don't need to end a line with a semicolon in shell scripting. The semicolon just exists to allow you to put two statements on one line.

Here's a cleaned up version of your script:

#!/bin/bash

VAR1=10
VAR2=20
if (( (VAR1 % 2 == 0) && (VAR2 % 5 == 0) )); then
    : #something (colon is a no-op)
fi