warning: string literal in condition

Using the first bit of code below I receive two warning messages:
`warning: string literal in condition` x2

if input == "N" || "n"
#do this
else input == "L" || "l"
#do this

as opposed to using this which results in no warnings

if input == "N" || input == "n"
#do this
else input == "L" || input == "l"
#do this

I'm wondering why the first bit of code results in a warning, and the downside of using it.

change `input == "N" || "n"`

to

input == "N" || input == "n"

You must also use `else if` instead of `else`


**The warning is saying that instead of a boolean or test, you have a string literal, ' n', which always evaluates to true.**

I was also looking for the answer to this question, and thanks to a friend found a few other solutions.

1) Change the case for the input so you only have to make one test:

if input.downcase == "n"

2) Use a more sophisticated check on the input data:

if %w{n N}.include?(input) or
if ['n', 'N'].include?(input)

The second one allows for far more flexibility with your checking , especially if there are groups of entry you are looking for.

Hope what I found helps others.

When you are writing `input == "N" || "n"`( internally Ruby sees it (`input == "N") || "n"`), it means `"n"` string object is always a truth value. Because in Ruby every object is `true`, except `nil` and `false`. Ruby interpreter is warned you there is not point to put ever true value is conditional checking. Conditional check statement always expect equality/un-equality test kind of expression. *Now you can go ahead this way or re-think again.* `if input == "N" || input == "n"` is not throwing any warning, as it obeys the norm of conditional test.

`else input == "L" || "l"` is wrong, as else statement don't expect any conditional test expression. Change it to `elseif input == "L" || "l"`

I have the same error as you but a different problem, found this through Google might as well post my solution for the next Googler.

I had a typo in my code which also gives the same warning:

if input =! "N"

of course the right way:

if input != "N"