I am writing a bash program that terminate program if entered number is over 100 and I used if [ $a -gt 100 ] (enter) exit 0 what wrong with this?

if [ $a -lt 7 ] || [ $a -gt 77 ]
then
echo "follow instructions"
elif [ $a -gt 100 ]
echo "invalid"
exit 0
fi 

but the exit statement isnt working

i was expecting the program to stop executing if I enter a number -gt 100

Two things:

• When a > 100 is true, a > 77 is also true.

• When an if clause is true, an elif is not even evaluated.

Consider instead:

#!/usr/bin/env bash
#              ^^^^- note, run this with bash, not sh

if (( a > 100 )); then
  echo "invalid" >&2
  exit 1
elif (( a < 7 || a > 77 )); then
  echo "follow instructions" >&2
fi

The conditional ranges do not make sense.

• You have (a < 7) or (a > 77) as the positive case.
• However, the (a > 77) positive case overlaps with (a > 100) negative case.
• Also, the program does not clarify what happens when (a >= 7 and a <= 77)?

Why would you want the program to exit on invalid input? Typically, a program validates input and asks repeatedly until it gets a value? You do not show the code where the user enters input. Below attempts to show both the user input with validation and tries to honor your completion and exit conditions:

#!/bin/bash
while [ 1 == 1 ];
do
  echo -n "Enter a number: "
  read a
  if ! [[ "$a" =~ ^[0-9]*$ ]]; then
    echo "Not a number, try again."
    continue
  fi
  if (( a > 100 )); then
    echo "Termination condition found."
    exit 1
  fi
  if (( a < 7 || a > 77 )); then
    echo "Valid number found."
    break
  fi
  echo "Invalid number found. Try again."
done
echo "You typed: $a"

Here are sample runs of the above program:

$ ./script.sh
Enter a number: afdafdsa
Not a number, try again.
Enter a number: fdaslkf;daslf;ds
Not a number, try again.
Enter a number: 55
Invalid number found. Try again.
Enter a number: 1234
Termination condition found.
$ ./script.sh
Enter a number: 3
Valid number found.
You typed: 3
$