if statement in javascript always true

Question

So, I have the code, its not done, but all i want it to do is display one alert box if I write the word \'help\', and say something else if anything else is entered.

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Answer 1

The problem is this line:

 if (reply === 'help' || 'Help')

Because in JavaScript, objects and non-empty strings evaluate to true when used as a boolean. There are a couple of exceptions to this when using ==

 if("0") // true
 if("0" == true) // false

In general, it's not a good idea to use == or raw variables in if statements.

As others have pointed out, use

if (reply === 'help' || reply === 'Help')

Or better:

if (typeof reply === 'string' && reply.toLowerCase() === 'help')

instead.

Answer 2

The reason why it always pops up is that reply === 'help' || 'Help' evaluates as (reply === 'Help') || ('Help'). The string literal Help is always truthy in Javascript hence it always evaluates to truthy.

To fix this you need to compare reply to both values

if (reply === 'help' || reply === 'Help') {
  ...
}

Or if you want any case variant of help use a regex

if (reply.match(/^help$/i)) {
  ...
}

Answer 3

The problem is here:

if (reply === 'help' || 'Help') // <-- 'Help' evaluates to TRUE
                                //      so condition is always TRUE

The equality operator doesn't "distribute", try

if (reply === 'help' || reply === 'Help')

Answer 4

if (reply === 'help' || 'Help')

should be:

if (reply === 'help' || reply === 'Help')

since 'Help' is "truthy" and so the first part of the if will always be entered.

Of course, even better would be to do a case-insensitive comparison:

if (reply.toLowerCase() === 'help')

Example: http://jsfiddle.net/qvEPe/

Answer 5

Just change this: if (reply === 'help' || 'Help')

To this: if (reply === 'help' || reply === 'Help')

The or statement was not comparing the variable.