How to check if integer value is less or more then 3

Question

We have two values

$a
$b

we need to compare the $a value with $b value

in case $b value is less than ($a - 3) or more than ($a + 3), then it will print fail.

example:

a=10
b=14

then it should fail.

For:

a=10
b=11

then it's ok.

For:

a=23
b=6

then it should fail.

You applied the awk tag. Do you want your solution in awk or will plain bash do? — Edward, Jul 14 '19 at 13:17
what you think its the best , depend to you ( even perl one liner ) — yael, Jul 14 '19 at 13:21
"in case $b value is less then 3 or more then 3 then it will print fail" so it should only pass if $b is exactly 3? where does the value of $a come in to the calculation? — steeldriver, Jul 14 '19 at 13:29
"The absolute value of their difference must be less than 3"? — Jeff Schaller, Jul 14 '19 at 13:40
Is your question (and this is a guess): ”How can I tell if the, numeric, values of two variable differ by more than three?” — ctrl-alt-delor, Jul 14 '19 at 17:07

score 1 · Accepted Answer · answered Jul 14 '19 at 13:29

1

I can't quite work out what exact numerical comparison you want to make, but in general in Bash arithmetic can be done as follows:

#!/bin/bash
a=100;
b=200;
threshold=50;

if [ $(($b - $a)) -gt $threshold ]
then
   echo Something.
else
   echo Something else.
fi

answered Jul 14 '19 at 13:29

Edward

FedKad · Answer 2 · 2019-07-14T19:49:16.327

0

Use bash arithmetic:

if (( (a-b) > 3 )) || (( (b-a) > 3 )); then
  echo fail
fi

Based on @ctrl-alt-delor's guess.

edited Jul 14 '19 at 19:49

answered Jul 14 '19 at 19:23

FedKad

2 Answers2