Conditional in shell script

Question

OS: ubuntu 14.04

I have the following script:

#!/bin/bash
read -rsp $'Press Y/y to execute script or any other key to exit...\n' -n1 key
if [ "$key" != 'y' ] || [ "$key" != 'Y' ] ; then
    echo 'You did not press the y key'
    exit
fi
echo 'some other stuff'

No matter what key I press, it always echoes "You did not press the y key"

What am I doing wrong?

anubhava · Accepted Answer · 2016-04-21 11:14:15Z

2

You need && instead of || as logic should say:

if key is not 'y' AND if key is not 'Y' then: error

Code:

if [ "$key" != 'y' ] && [ "$key" != 'Y' ] ; then
    echo 'You did not press the y key'
    exit
fi

If you're using bash then you can shorten that to:

if [[ $key != [yY] ]]; then
    echo 'You did not press the y key'
    exit
fi

answered Apr 21, 2016 at 11:14

anubhava

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2 Comments

EastsideDev Over a year ago

Duh, yes. Too early in the morning. Is the [[]] necessary for the shortened code, or can I use only one set of brackets?

Etan Reisner Over a year ago

@EastsideDeveloper You need [[...]] to use a pattern as the RHS of !=.

ghoti · Accepted Answer · 2016-04-21 20:15:15Z

2

What anubhava said. Also, case might be a good alternative to your if statement:

#!/usr/bin/env bash

read -rsp $'Press Y/y to execute script or any other key to exit...\n' -n1 key

case "$key" in
  Y|y)
    echo 'You DID press the y key. Proceeding.'
    ;;
  *)
    printf "I can't handle the '%s' key!\n" "$key"
    exit 1
    ;;
esac

echo 'some other stuff'

Incidentally, except for the read options, this is POSIX-compatible, so it's more portable than bash-only code (like anything using [[ ... ]] conditions).

edited Apr 21, 2016 at 20:15

answered Apr 21, 2016 at 11:28

ghoti

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