Two conditions in a bash if statement

Question

I\'m trying to write a script which will read two choices, and if both of them are \"y\" I want it to say \"Test Done!\" and if one or both of them isn\'t \"y\" I want it to

Answer 1

The program is doing exactly what you told it to do. You said "If the first choice is not equal to 'y' and the second choice is not equal to 'y' then print "Test Done !" otherwise print "Test Failed !" -- so only if both choices are not y will "Test Done !" be printed.

You probably meant:

echo "- Do You want to make a choice ?"
read choice

echo "- Do You want to make a choice1 ?"
read choice1

if [ "$choice" == 'y' ] && [ "$choice1" == 'y' ]; then
echo "Test Done !"
else
echo "Test Failed !"
fi

I changed != not equals to == equals. Now only if you answer "y" to both questions will "Test Done !" be printed.

Answer 2

You got the comparison logic backwards; from your description you wanted to say

if [ "$choice" = 'y' ] && [ "$choice1" = 'y' ]; then

I'm actually surprised that the && construct works, although on further inspection it probably should. Still, I would write it as

if [ "$choice" = 'y' -a "$choice1" = 'y' ]; then