How can I use a while loop to execute this code properly in UNIX?

Question

echo -n "Enter a positive integer: "; read integer
  While [ $integer -gt 0 ]; do
    echo "$integer"
  done

I am trying to write a script in UNIX that meets the following criteria:

Name the script while.sh.
Ask the user to enter a positive integer. You can assume that the user will enter a positive integer (input validation not required).
Use a while loop to print all integers from 0 up to and including the integer entered.

The first two steps were easy, but I cannot get the third step to execute properly. Can anyone help me?

You need a "counter" variable that starts at 0 and stops when greater than the user's value. — glenn jackman
– glenn jackman, Commented Apr 7, 2015 at 19:09
I did not know the expr command, that will be useful in the future, thanks for your input. — Dresden Feitzinger
– Dresden Feitzinger, Commented Apr 7, 2015 at 19:32
Don't use expr for integer arithmetic; use the POSIX arithmetic expression; x=$((x + 1)). — chepner
– chepner, Commented May 2, 2015 at 2:39

Breno Leitão · Accepted Answer · 2015-04-07 19:17:32Z

2

This is the script you want:

#!/bin/bash

echo -n "Enter a positive integer: "; read integer
while [ $integer -gt 0 ]; do
    echo "$integer"
    integer=`expr $integer - 1`
done

This is an example:

./while.sh 
Enter a positive integer: 10
10
9
8
7
6
5
4
3
2
1

answered Apr 7, 2015 at 19:17

Breno Leitão

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2 Comments

bobah Over a year ago

or simply ((--integer))

chepner Over a year ago

((--integer)) is not a standard shell command; only some shells will support it. Instead, use integer=$((integer - 1)) or even : $((--integer)).

biggie747 · Accepted Answer · 2015-04-07 19:35:54Z

0

This is the script I came up with. I may have taken this a little to literal, however your question does ask for displaying 0 to the number entered.

#! /bin/bash

i=0

echo -n "Enter a positive integer: "
read integer

while [ $integer -ge $i ]
 do
  echo "$i"
  ((i++))
done

Here is the output from above:

Enter a positive integer: 11
0
1
2
3
4
5
6
7
8
9
10
11

answered Apr 7, 2015 at 19:35

biggie747

1

1 Comment

Dresden Feitzinger Over a year ago

That is another way to think about it. I like your way because it goes in ascending order, rather than descending.

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