Shell Script Too Many Arguments for if condition

Question

My current script does the following;

It takes integer as a command line argument and starts from 1 to N , it checks whether the numbers are divisible by 3, 5 or both of them. It simply prints out Uc for 3, Bes for 5 and UcBes for 3,5. If the command line argument is empty, it does the same operation but the loop goes to 1 to 20.

I am having this error "Too many arguments at line 11,15 and 19".

Here is the code:

#!/bin/bash

if [ ! -z $1 ]; then
    for i in `seq 1 $1`
    do
        if [ [$i % 3] -eq 0 ]; then
            echo "Uc"
        elif [ i % 5 -eq 0 ]; then
            echo "Bes"
        elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
        then
            echo "UcBes"
        else
            echo "$i"
        fi
    done
elif [ -z $1 ]
then
    for i in {1..20}
    do
        if [ i % 3 -eq 0 ]
        then
            echo "Uc"
        elif [ i % 5 -eq 0 ]
        then
            echo "Bes"
        elif [ i % 3 -eq 0 ] && [ i % 5 -eq 0 ]
        then
            echo "UcBes"
        else
            echo "$i"
        fi
    done
else
    echo "heheheh"
fi

@[A. Mesult Konuklar] I edited your question to rollback to previous edit: you may not remove the error from the initial question, otherwise you'll make the overall post unusable to others. Hope you got it by now, but just to explain. — J. Chomel
– J. Chomel, Commented Apr 30, 2020 at 5:50

sampson-chen · Accepted Answer · 2012-11-15 21:16:03Z

6

Note that [ is actually synonym for the test builtin in shell (try which [ in your terminal), and not a conditional syntax like other languages, so you cannot do:

if [ [$i % 3] -eq 0 ]; then

Moreover, always make sure that there is at least one space between [, ], and the variables that comprise the logical condition check in between them.

The syntax for evaluating an expression such as modulo is enclosure by $((...)), and the variable names inside need not be prefixed by $:

remainder=$((i % 3))
if [ $remainder -eq 0 ]; then

edited Nov 15, 2012 at 21:16

answered Nov 15, 2012 at 20:49

sampson-chen

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11 Comments

A. Mesut Konuklar Over a year ago

Thank you, I just leanred [ meant test in shell. I put space between them normally , but tried without space also :)

Julien Vivenot Over a year ago

You say variables should not be prefixed within $((...)), but both $(($i%3)) and $((i%3)) seem to return the same.

gniourf_gniourf Over a year ago

In bash, [ is a builtin, as shown by help [, and not as written here a link to the test command.

sampson-chen Over a year ago

@gniourf_gniourf help [ prints This is a synonym for the "test" builtin, at least in bash.

gniourf_gniourf Over a year ago

Yes, you read properly: builtin !!! It's not the same as the command [ and test. which looks for commands in your path...

|

Julien Vivenot · Accepted Answer · 2012-11-15 20:50:15Z

3

You should probably use something like :

if [ $(($i % 3)) -eq 0 ]

instead of

if [ $i % 3 -eq 0 ]
if [ [$i % 3] -eq 0 ]

answered Nov 15, 2012 at 20:50

Julien Vivenot

2,25013 silver badges17 bronze badges

2 Comments

A. Mesut Konuklar Over a year ago

Thank you, I though I've tried this method also, but I did it like $((i%3)) silly me. :)

Julien Vivenot Over a year ago

Actually, as @sampson-chen pointed out in his solution, $((i%3)) works.

William Pursell · Accepted Answer · 2012-11-15 22:29:27Z

1

Your script could be greatly simplified. For example:

#!/bin/sh

n=0
while test $(( ++n )) -le ${1:-20}; do
  t=$n
  expr $n % 3 > /dev/null || { printf Uc; t=; }
  expr $n % 5 > /dev/null || { printf Bes; t=; }
  echo $t
done

gives slightly different error messages if the argument is not an integer, but otherwise behaves the same.

answered Nov 15, 2012 at 22:29

William Pursell

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