conditional binary operator expected , what is the problem? (shell)

Question

I was trying to write a shell-script that would print all the arguments passed to the script. The condition is, the argument has to either be an uneven number or smaller than 23

#!/bin/bash
for (( i=0; i<$#; i++ ));
do
        if [[ $i % 2 -ne 0 ]] || [[ $i -lt 23 ]];
        then
                echo $i
        fi
done

When I try to Run this script as followed sh myScript.sh 1 3 4 6 24 23 22 My expected output would be 1 , 3 , 4 , 6 , 22 , 23

However I get the following Error:

myScript.sh: line 4: conditional binary operator expected
myScript.sh: line 4: syntax error near `%'
myScript.sh: line 4: `        if [[ $i % 2 -ne 0 ]] || [[ $i -lt 23 ]];'

Can anyone help me understand what the error means?

Thank you!

% is not an operator in [[...]]. Use if ((i % 2 != 0 || i < 23)). — M. Nejat Aydin
– M. Nejat Aydin, Commented Jul 7, 2020 at 20:10
@M.NejatAydin Thank you. I tried your suggestion, but I get a different Error now: myScript.sh: line 5: syntax error near unexpected token `then' myScript.sh: line 5: ` then' — sepæpa
– sepæpa, Commented Jul 7, 2020 at 20:14
Most likely you forgot a semicolon. if ((i % 2 != 0 || i < 23)); then echo $i; fi should work. — M. Nejat Aydin
– M. Nejat Aydin, Commented Jul 7, 2020 at 20:16

Ted Lyngmo · Accepted Answer · 2020-07-07 20:21:12Z

1

With Arithmetic Expansion you can do it like this:

#!/bin/bash

for var in "$@"
do
    if (( var%2 || var<23 ))       # arithmetic expansion
    then
        echo $var
    fi
done

Output:

edited Jul 7, 2020 at 20:21

answered Jul 7, 2020 at 20:10

Ted Lyngmo

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3 Comments

chepner Over a year ago

It should be checking if argv[i] (not i) is odd or less than 23.

Ted Lyngmo Over a year ago

@chepner Ok, I must have read his code wrong.... Oh yeah, fixed it.

sepæpa Over a year ago

@TedLyngmo Thank you for your Help! This works fine, and you showed me something new "Arithmetic Expansion".

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