How to find a multiple of a number within an array in bash?

Question

I am writing a bash script which runs through the numbers 1 - 50 and I have to output every number except numbers which are multiples of 4 (ex. 4, 8, 12..). I have tried using an example from a similar question which asks to output the same array except certain numbers.

In the code attached my program will output every number except 3.

    #!/bin/bash
LIMIT=49
echo "Printing multiples of 4 from 1 - 50: "
a=0
while [ $a -le $LIMIT ];do
    a=$(($a+1))
    if [$a -eq 3]
    then 
        continue
    fi
    echo -n "$a"
done

How should I change the IF statement to output my desired script?

Why not just seq 49 | sed '4~4d'?

oguz ismail
– oguz ismail

2020-04-02 13:11:49 +00:00
Commented Apr 2, 2020 at 13:11 — oguz ismail
– oguz ismail, Commented Apr 2, 2020 at 13:11

oguz ismail · Accepted Answer · 2020-04-02 13:49:38Z

2

for num in {0..50}; do
  if (( num % 4 )); then 
    echo $num
  fi
done

This is not as brief as sed version, but it shows how to achieve it with bash built-ins only.

edited Apr 2, 2020 at 13:49

oguz ismail

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answered Apr 2, 2020 at 13:11

vtronko

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2 Comments

user12978906 Over a year ago

Although this is more or less on the right track, how would I change the code if I wanted the output to 'not' be multiples of 4? Ex: 1,2,3,5,6,7,9

vtronko Over a year ago

@MiSully huh? This is exactly what it outputs, non multiples of 4.

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