CODEWITHSUNDEEP
bash
Connor Cartwright
Connor Cartwright

Reputation: 385

Bash; conditional statement echoing numbers

I'm nearly done writing a script for an assignment, but am having some trouble thinking of how to do this final part.

My problem is within a while loop; it prints out the number based on the IF statements, the number entered will always be an even number.

The IFs aren't connected by else/elif because the number should be able to printed out if it applies to more than 1 of the statements.

I want to print $starting on every loop if it doesn't meet any of the IF conditions, but if it does I don't want to print it. Can anyone see how to do that?

while [[ $starting -lt $ending ]]; do
    if [ $((starting %7)) -eq 0 ] 
    then
    echo "$starting red"
    fi
    if [ $((starting % 11)) -eq 0 ]
    then
    echo "$starting green"
    fi
    if [ $((starting % 13)) -eq 0 ]
    then
    echo "$starting blue"
    fi
    starting=$((starting + 2))
done

Upvotes: 2

Views: 72

Answers (2)

that other guy
that other guy

Reputation: 123490

Keep track of whether you've done what you want to do in a variable:

while [[ $starting -lt $ending ]]; do

handled=0

if [ $((starting %7)) -eq 0 ] 
then
echo "$starting red"
handled=1
fi
if [ $((starting % 11)) -eq 0 ]
then
echo "$starting green"
handled=1
fi
if [ $((starting % 13)) -eq 0 ]
then
echo "$starting blue"
handled=1
fi

if ! (( handled ))
then 
  echo "$starting didn't match anything"
fi

starting=$((starting + 2))
done

Upvotes: 3

zbrunson
zbrunson

Reputation: 1757

Add another if at the end that checks if none of the previous if statements are true. if !(starting%7==0 or starting%11==0 or starting%13==0) => echo starting

Upvotes: 2

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