UNIX for Dummies Questions & Answers

I am receiving an elif error on line 13 and I can not figure out the reasoning behind it. I have added the then statement that I was initially missing. Any help would be great.


#The purpose of this script is for the end user to be able to enter a positive number
#User enters a number
NUM=$1
#If user enters more than one it will return this statement
if [ $# -ne1 1 ] then
echo "Only enter one number!"
exit
#For if the user enters a negative number
elif [ $NUM -le 0 ] then
echo "Enter a positive number!"
fi
#Executionary if statement
while [ "$NUM" -gt 0 ]
do
printf $NUM
if[ $NUM -gt 1 ] then
printf ","
fi
NUM=$((NUM - 1))
done


Thanks

You are missing a semicolon before an earlier then. Also -ne is misspelled.

The syntax of the if statement is

Code:

if some command or other; then
   action
else
   otheraction
fi

Because "some command or other" can be as long or short as necessary, you really do need a semicolon or a newline before the "then".

It still shows me as having an issue with the done portion of my while loop at the end, I tried moving the done above and below the last statement but it wont work.

Did you fix the if inside the while also?

Nope...that fixed it though.

Thanks.