How to while loop correctly in bash script?

Question

I have wrote this script which takes a stop and start number and counts out the numbers in between - I am yting to get it to keep increasing no matter whether the "stop" number is reaced or not so it keeps counting until ctrl+z is pressed but it is not recognising the while condition for me - could anyone correct the syntax for me please ?

  #!/bin/sh

stopvalue=$1
startvalue=$2

if [ $# -ne 2  ]
then
   echo "Error - You must enter two numbers exactly - using default start value of 1"
#exit 0
fi

echo ${startvalue:=1}

while (test "$startvalue" -le "$stopvalue" ||  "$startvalue" -ge "$stopvalue")
do
       startvalue=$((startvalue+1))
        echo $startvalue
done

Michael Krelin - hacker · Accepted Answer · 2011-12-05 19:08:53Z

14

Now that you have two answers about the while loop, I'll suggest using for loop instead:

for((i=$startvalue;i<=$stopvalue;++i)) do
   echo $i
done

answered Dec 5, 2011 at 19:08

Michael Krelin - hacker

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Comments

Tarun Sharma · Accepted Answer · 2015-10-14 16:18:33Z

5

I use the simple while loop program. Which works for me. as per my understanding before passing any variable in while loop we have to declare it:

sum=0            
temp=1          
while [ $temp -ne 0 ]           
do              
echo "Enter number: OR 0 to Quit :"             
read temp          
sum=`expr $sum + $temp`          
done          
echo $sum

answered Oct 14, 2015 at 16:18

Tarun Sharma

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Comments

Community · Accepted Answer · 2021-03-22 12:24:13Z

5

Your main problem is in your condition:

while (test "$startvalue" -le "$stopvalue" ||  "$startvalue" -ge "$stopvalue")

You are telling it to continue if $startvalue is (less than or equal to) or (greater than or equal to) $stopvalue. For ANY combination of $startvalue and $stopvalue, the one will be less than, equal to, or greater than the other. Take out the second condition so that it only continues if $startvalue is less than or equal to $stopvalue.

And you should write it:

while [ "$startvalue" -le "$stopvalue" ]

edited Mar 22, 2021 at 12:24

CommunityBot

11 silver badge

answered Dec 5, 2011 at 19:02

Kevin

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Comments

Andrew Clark · Accepted Answer · 2011-12-05 19:00:33Z

0

while [ "$startvalue" -le "$stopvalue" -o "$startvalue" -ge "$stopvalue" ]
do
       startvalue=$((startvalue+1))
       echo $startvalue
done

answered Dec 5, 2011 at 19:00

Andrew Clark

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4 Comments

Kevin Over a year ago

That's still going to loop forever because $startvalue is always less than (first condition), equal to (both), or greater than (second) $stopvalue.

Andrew Clark Over a year ago

@Kevin - Good point, I didn't spend much time looking at whether or not the loop made sense, I was just fixing the syntax.

Cougar Over a year ago

This is correct syntax. In test you have to use -o and not ||. || is totally different thing and after that you need another shell command (like test).

Michael Krelin - hacker Over a year ago

you can use || with [[ ]], like [[ condition || condition ]]

Ifeanyi · Accepted Answer · 2021-10-13 12:48:20Z

-2

startvalue=50
stopvalue=0
while [ $startvalue -lt  $stopvalue -o $startvalue -ne $stopvalue ]; do
echo "Enter number equal to the start value: :"             
read stopvalue
done

answered Oct 13, 2021 at 12:48

Ifeanyi

1

1 Comment

Stephen Ostermiller Over a year ago

A code-only answer is not high quality. While this code may be useful, you can improve it by saying why it works, how it works, when it should be used, and what its limitations are. Please edit your answer to include explanation and link to relevant documentation.

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