Shell Script Too Many Arguments for if condition
bash, shell, terminal
Solution
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
Problem
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 ```