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

Try:

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

Answer 2

Another thought,

$ c1='y' ; c2='y' ; [[ ${c1} = 'y' ]] && [[ ${c2} = 'y' ]] && echo true || echo false  
true  
$ c1='n' ; c2='y' ; [[ ${c1} = 'y' ]] && [[ ${c2} = 'y' ]] && echo true || echo false  
false  
$ c1='n' ; c2='y' ; [[ ${c1} = 'y' ]] || [[ ${c2} = 'y' ]] && echo true || echo false  
true  
$ c1='n' ; c2='n' ; [[ ${c1} = 'y' ]] || [[ ${c2} = 'y' ]] && echo true || echo false  
false  
$  

Overflow of gibberish. (;

Answer 3

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

Answer 4

You have your logic reversed; you're checking for != when you should be checking for ==. Try this:

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

Answer 5

You are checking for the wrong condition.

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

The above statement is true when choice!='y' and choice1!='y', and so the program correctly prints "Test Done!".

The corrected script is

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

Answer 6

The line

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

tests if both choices aren't 'y', so when both choices are 'y', the statement is false and your program correctly prints "Test Failed".