shell script - while loop does not terminate

Question

I have a while loop in shell and it does not terminate even though it has a break condition, this piece of code run well in other languages but i can't figure out how while loop runs in shell script. Here is my code:

NAME=""
while [[ $NAME != "q"  || $NAME != "Q" ]]
do
    read NAME
    //do something


done

NAME is always not "q" or not "Q". For example, "q" is not "Q"; "Q" is not "q"; and "x" is neither "q" nor "Q". You want &&, not ||. (This would behave identically in other languages, the error is purely logical.) — Amadan
– Amadan, Commented Mar 5, 2019 at 2:14
@Amadan yes thank you very much, code in shell looks different from others, i get it now. — Thuan Nguyen
– Thuan Nguyen, Commented Mar 5, 2019 at 2:16

paxdiablo · Accepted Answer · 2019-03-05 02:22:10Z

1

$NAME != "q" || $NAME != "Q"

Unless NAME can exist in a strange " Schrödinger's cat" sort of state where it can be both Q and q at the same time, this expression will always be true.

Think about it:

If NAME is neither q nor Q, both sub-expressions will be true so the full expression will be true.
If it's Q, then the first sub-expressions will be true, leading to the full expression being true as well.
If it's q, then the second sub-expressions will be true, leading to the full expression being true as well.

What you probably need is:

$NAME != "q" && $NAME != "Q"

Of course, if it's bash that you're using, it provides a way to uppercase and lowercase a string to make these comparisons easier:

while [[ "${NAME^^}" != "Q" ]] ; # upper-case variant
while [[ "${NAME,,}" != "q" ]] ; # lower-case variant

edited Mar 5, 2019 at 2:22

answered Mar 5, 2019 at 2:16

paxdiablo

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1 Comment

Benjamin W. Over a year ago

Or you could use [[ $NAME != [Qq] ]]

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