While loop not equal command not found

Question

I am trying to perform a simple while loop comparing a variable to a string. It fails to load with the error on this line. Error states [: missing `]' and : command not found. My while loop looks like;

mainMenuInput = ""
while ["$mainMenuInput" != "Q" || "$mainMenuInput" != "q"]
do

fedorqui · Accepted Answer · 2014-03-25 11:08:26Z

2

There are a couple of errors:

mainMenuInput="" 
while [ "$mainMenuInput" != "Q" ] && [ "$mainMenuInput" != "q" ]

See that variable declaration do need to be like var=value. Otherwise, bash would interpret that you want to perform the var command with = and value as parameters:

mainMenuInput="" 
             ^
             no spaces around =

In the while you need to put spaces around the brackets. Also, note that you need to use && (and) instead of || (or), because otherwise it won't ever exit the while.

while [ "$mainMenuInput" != "Q" ] && [ "$mainMenuInput" != "q" ]
       ^                          ^^                          ^
   space                   it has to be AND               space

In fact, the condition can be rewritten to something like:

while [[ "$mainMenuInput" != [qQ] ]]

edited Mar 25, 2014 at 11:08

answered Mar 25, 2014 at 10:46

fedorqui

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4 Comments

user1046334 Over a year ago

plus -o instead || or -a instead && when used inside []

fedorqui Over a year ago

Thanks, @herby but POSIX recommends the use of && and || with the single bracket notation over -a and -o.

user2966161 Over a year ago

Thanks a lot guys, I didn't know you had to have no spaces around the '=' thanks. Does what i needed it to now thanks. Also @herby using -o did work for displaying the contents for the do thanks.

fedorqui Over a year ago

@Seatter variables are set with var=value expression, without any space around =. Otherwise, bash would interpret that you want to perform the var command with = and value parameters.

Arif Burhan · Accepted Answer · 2016-02-25 04:29:41Z

0

If you are using bash I think the correct syntax is:

    while [ $variable -ne "q" || $variable -ne "Q" ];

answered Feb 25, 2016 at 4:29

Arif Burhan

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