Score:0

Ubuntu

Bash from scratch

Sneakatoo

10/15/22, 10:57 AM

I know this may sound ridiculous, but in my defence I'll say I'm at the very beginning of messing around with coding and bash in particular, so don't judge me too harshly...

I just dipped into this:

#!/bin/bash

echo -n "Enter a number: "
read VAR

if [[ $VAR -gt 10 ]]
then
  echo "The variable is greater than 10."
fi

Thing is any number I write up doesn't change a thing. Even 2 will be greater than 10. What's wrong?

2 + 3

command-line

bash

scripts

Karlom

10/15/22, 11:05 AM

This scrpt works fine for me: `Enter a number: 12 The variable is greater than 10.` What's exactly the problem ?

paladin

10/15/22, 11:18 AM

Please do `ls /bin/bash -l` and tell us what the output is. `/bin/bash --version` would also help.

terdon

10/15/22, 5:25 PM

You have two bad practices there: i) avoid using CAPS for variables. By convention, global environment variables are in CAPS and if you set yours that way too that can cause surprising bugs. ii) Always quote your variables when using them.

Score:5

Ubuntu

Artur Meinild

10/15/22, 11:05 AM

The script has a correct syntax, and works for me. If I enter anything from 0-10, it does not echo any text.

For future reference, I recommend you install shellcheck and make a habit of checking your script syntax with it.

0 + 0

Score:2

Ubuntu

zwets

10/16/22, 9:10 PM

Alternatively (since you mention that you are learning bash) you could use an arithmetic expression:

if (( VAR > 10 ))
then
    echo "The value is greater than 10."
fi

Or the more concise one-liner:

(( VAR <= 10 )) || echo "The value is greater than 10."

0 + 0

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Bash from scratch

TH: ทุบตีตั้งแต่เริ่มต้น

RO: Bash de la zero

RU: Баш с нуля

VI: Bash từ đầu

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