entering charcter or numerical value then checking

[lcpdl87]

Write a shell script to find even or odd, and print out the result.

This should include 2 portions:

Ask user to input a numerical number, then the script should check if it is a character or numerical, ask user re-enter a numerical number, if character found.

If it is an numerical digit, find out even or odd, and print out the result


this is what I got:


#!bin/bash
echo -e "Please enter a numerical number --> \c"
read NUMBER

mychar=1
if echo -e "$mychar | grep [a-z]"
then
echo -e "Please re-enter a numerical number --> \c"
read NUMBER

if [ $(($number % 2)) -eq 0 ]
then
echo -e "$number is even even number"
else
echo -e "$number is an odd number"
fi


... am i doing something wrong. I'm assuming so. Any help would be awesome!

[Mel Burslan]

your script is missing some stuff.

do you want the user to be forced to enter a numeric value or do you want him to exit the script with a non-zero return value, if he/she doesn't enter a numeric ?

depending on the answer, you either will need an endless loop to check until a numeric value was entered around the first if block, something like:

numeric_flag=0
while [ ${numeric_flag} -eq 0 ]
do
# check for alphabetic characters here
# if numeric value entered set numeric_flag to 1
done

otherwise,

if non numeric value was found

exit 1

This should be a starting point for your debugging

[lcpdl87]

I want the user to enter a "numerical number" like it says in the directions and if they enter an "a-z" character like a charcater in the alphabetic, then it should go down to "please re-enter another number" and then if someone enters 0-9.. it reads whether the number is odd or even.

And obviously if the person enters a "0-9" on the first try, the programs read sthat and figures out if its odd or even right away.

[Mel Burslan]

actually, let me re-phrase it, it has a lot of mistakes, like variables being assigned in upper case whereas getting checked in lower case. You are not using a mainframe or DOS or an archaic high level language which is case insensitive. In unix $NUMBER is not equal to $number which is not equal to $NuMbEr.

here is the script you are looking for:

r=0
while [ $r -eq 0 ]
do
clear
echo "enter numeric value"
read num
echo $num | grep -q [a-z]
r=${?}
done

result=$(($num % 2))
if [ $result -eq 0 ]
then
echo "\n\nthe number you entered was $num and it is an even number\n\n"
else
echo "\n\nthe number you entered was $num and it is an odd number\n\n"
fi

[lcpdl87]

well,

the "result=$(($num %2))
if[$result -eq 0]

is giving me an error..

syntax error: operand expected (error token is "%2")
then..
if [-eq 0]: command not found
syntax error near unexpected then

[Steven Schweda]

> Write a shell script to find even or odd,
> and print out the result.

Sure sounds like homework to me.

[lcpdl87]

No, its extra credit, and I believe I have most of it now but I'm still getting errors.

[Mel Burslan]

extra credit is still homework or course study of some sort and instead of copying someone else's work, you should be doing your own investigation. If I were to know/think about this, I would not be writing the script for you previously. I know the script above works on a hpux 11.11 machine. I can see where you are going wrong but will not tell you :) It would otherwise take the whole fun out.

Good luck

[lcpdl87]

Well, if I wasn't stuck, I wouldn't even bother checking this site for help. It's a forum, you post your problem and people help you out. That's how a forum works doesn't it? And I've been looking through the book I was given and on the internet and I'm not finding any accurate results that could help me out in the slightest bit. All I wanted was some guidance.

[lcpdl87]

And I'm working with Linux 10, not whatever you're talking about.