I just started programming on UNIX this week and one of my homework task is to create a script that reads positive numbers until -99 is entered and negative numbers show an error ( 0 to -98). This is what I've come up with but kinda stuck. thanks :)
#!/bin/bash
COUNTER=1
echo "Enter a positive integer number (-99 to quit):"
read NUMBER
echo "$NUMBER: "
while [ $NUMBER -ge $COUNTER ]; do
echo $COUNTER
((COUNTER++))
done
#!/bin/bash
while true; do
read -p 'Number (-99 to quit): '
if ! [[ "$REPLY" =~ ^-?[0-9]+$ ]]; then echo 'Error: Not an integer' >&2
elif (( REPLY == -99 )); then break
elif (( REPLY <= 0 )); then echo 'Error: Need positive integers >0' >&2
else
printf 'Got "%d", that is number %d\n' "$REPLY" "$(( ++c ))"
fi
done
This is an infinite loop that is exited when the user enters -99. Any positive integer response will prompt the code to say Got "some number" followed by how many valid numbers read so far, whereas a negative integer, zero, or a non-numeric input will give a diagnostic message on standard error. The code uses the variable REPLY which is the variable written to by read if not given any other variable name.
The test for correct numeric input is done with matching the response against the regular expression ^-?[0-9]+$. This expression will match if the response is on the form that we expect (an optional dash followed by at least one digit). If it doesn't match, a diagnostic message is issued on standard error.
Up until the first elif we can not be sure that $REPLY is an integer. After that, we use (( ... )) for arithmetic evaluation of the comparisons.
The test for -99 needs to come before the test for negative integers, as we otherwise would not have any way of exiting the loop.
then if you wish. I just wanted it to be compact. I wouldn't know Python if it jumped up and bit me in the nose.
If I understood you correctly, this should work:
#! /bin/bash -
read -p "Enter a positive integer (-99 to quit): " USR_INT
while [[ -n "$USR_INT" ]]; do
case "$USR_INT" in
-99)
echo "Exiting..."
exit 0
;;
-*)
echo "Error, please enter a positive integer."
exit 1
;;
0)
echo "Error, please enter a positive integer."
exit 1
;;
[0-9]*)
echo "You have entered $USR_INT"
;;
*)
echo "Error, please enter a positive integer."
exit 1
;;
esac
read -p "Enter a positive integer (-99 to quit): " USR_INT
done
Exit 1 line under the -*) part, however you said a negative number should generate an error.
Keeping it simple:
!/bin/bash
while :
do
echo "Enter a positive integer number (-99 to quit):"
read NUMBER
if (( NUMBER == -99 ))
then
exit
fi
echo "$NUMBER: "
COUNTER=1
while (( COUNTER <= NUMBER ))
do
echo $COUNTER
((COUNTER++))
done
done
COUNTERfor then? That is not explained by this assignment. Do the positive numbers have to be in order? Can they not be entered twice...with yourCOUNTERif a person entered5it would work and then if they entered2it would not, but2is a valid positive integer no?