while loop integer expression expected

Question

I have a problem with this while loop. I want the user to pick one of the three options. I get the error "integer expression expected". When i remove the last option "$UCHOICE" -ne "q", the program works until i type a letter, and i get the same "integer expression expected". I am guessing that the problem has to do with integers and strings but i have no clue. please help.

while [ "$UCHOICE" -ne "1" -a "$UCHOICE" -ne "2" -a "$UCHOICE" -ne "q" ]
do
echo "hello"
read UCHOICE
done

zzevannn · Accepted Answer · 2015-11-20 00:38:19Z

3

You need to be using != for string comparison.

while [ "$UCHOICE" != "1" -a "$UCHOICE" != "2" -a "$UCHOICE" != "q" ]
do
echo "hello"
read UCHOICE
done

edited Nov 20, 2015 at 0:38

answered Nov 20, 2015 at 0:32

zzevannn

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Comments

icecity96 · Accepted Answer · 2015-11-20 01:20:11Z

1

I think the following message will help you:

stringA != stringB stringA does not match stringB

exprA -ne exprB Arithmetic expressions exprA and exprB are not equal

answered Nov 20, 2015 at 1:20

icecity96

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