Get user input and check it, linux

Question

I am a beginner at unix and I am trying to use a while loop to get a user integer input for 2 numbers, but I need to check if it is an integer and reask if it's not, so I attempted to utilize if statements inside the while loop. I cannot get this to work, what am I doing wrong with the while and if loop? #! /bin/bash

    echo “Enter first number:“

    while read number1
    do
      if[[ $number1 ]] && [ $input -eq $input 2>/dev/null ]
      then 
        echo “$number1”

      else
        echo "$number1 is not an integer or not defined.Try again”
      fi
     done


    echo “Enter second number:“
    while read number2
       do
        if[[ $number2 ]] && [ $input -eq $input 2>/dev/null ]
            then 
              echo “$number2”
        else
              echo "$number2 is not an integer or not defined.Try again”
         fi
    done

Please include in your question the mandatory four: ① What did you try, ② what did you expect to happen, ③ what did you observe instead, and ④ why is this not matching your expectation. You provided ① and ②. — Alfe
– Alfe, Commented Apr 7, 2020 at 18:36
Check your spaces (always around [[ and ]]) and your quotes. — Walter A
– Walter A, Commented Apr 7, 2020 at 20:50

Vikash · Accepted Answer · 2020-04-08 06:56:38Z

1

If you are trying to get the number and checking it until it becomes integer, Please try the below code.

#!/bin/bash
echo "enter number"
while read number1
do
if ! [[ $number1 =~ ^[0-9]+$ ]]
then
        echo "number is not an integer"

else
        echo "number is an integer"
        exit;
fi
done

answered Apr 8, 2020 at 6:56

Vikash

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