now my code looks like this
!/bin/bash
read INPUT
for number in $(seq 1 $INPUT); do
if [ $((number%2)) -eq 0 ]
else [ $((number%2)) -eq 1 ]; then
echo $number
fi
done
but then I get the next error
./script.sh: line 6: syntax error near unexpected token `else'
./script.sh: line 6: ` else [ $((number%2)) -eq 1 ]; then'
Aucun commentaire:
Enregistrer un commentaire