Syntax for OR condition in bash while loop

Question

I tried many different combinations with brakets, quotation marks, ||, -o, but the only way the script works without infinit loop is without the OR in while loop comparison, like this: while [ $name != Jorge ]] ; do ... This is an example of an script that I want to run:

#!/bin/bash
echo "What is my name: "
read name

while [ $name != "Jorge" ] || [ $name != "Eduardo" ] ; do    
    echo "Not. Try again: "
    read name
done

echo "Well done!"

Since you are using bash, you could use the regex operator also while [[ ! $name =~ ^(Jorge|Eduardo)$ ]]; do — Inian
– Inian, Commented Aug 13, 2017 at 6:32

Mustafa Hussain · Accepted Answer · 2017-08-13 06:25:52Z

7

You should use && instead of || in the condition of while statement.

You are trying to read a name from stdin and if that name is "Jorge" or "Eduardo", you are done. when putting it in condition of while, you want to continue in the loop when the name if not "Jorge" and the name is not "Eduardo".

your current condition says that continue in the loop if the name is not "Jorge" or the name is not "Eduardo". And the name cannot be both at the same time.

answered Aug 13, 2017 at 6:25

Mustafa Hussain

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2 Comments

Leponzo Over a year ago

De Morgan's laws

Joe Over a year ago

Would not have realized I did this too, if you hadn't explained it so well. Thank you!

SLePort · Accepted Answer · 2017-08-13 16:23:56Z

2

Using regex:

while [[ ! "$name" =~ ^(Jorge|Eduardo)$ ]]; do

or Bash extented globbing:

shopt -s extglob
while [[ "$name" != @(Jorge|Eduardo) ]]; do

edited Aug 13, 2017 at 16:23

answered Aug 13, 2017 at 6:26

SLePort

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2 Comments

chepner Over a year ago

-a is obsolete and should not be used, per the POSIX specification for test.

SLePort Over a year ago

@chepner Removed from my answer.

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