Bug in jQuery 1.9.1 with function $.inArray

Question

I recently came to conclusion that $.inArray is not working. I researched everywhere, saw similar threads to jQuery: why doesn't jQuery.inArray() work?

But still I see it as a jQuery Bug. Because I have a ARRAY which I am using in $.inArray

Here is the console.log

var testArray value is:

["1", "15"]

which is obvisouly an array in JavaScript.

And I am using command

if($.inArray(14, testArray))
  alert("14 should not be in array. Isn't it?");

I Also tried following, but stil same problem:

if(jQuery.inArray('14', testArray))
  alert("14 should not be in array. Isn't it?");

Answer 1

jQuery's $.inArray method returns an index of where the value is found, or -1 if it is not found. It just so happens that -1 is truthy (convert it to a boolean by running !!-1 ), which will evaluate to true. In order to more reliably use this method, compare its return value rather than merely using it as a condition:

if ( $.inArray(14, ["1", "15"]) > -1 ) {
    alert( "Value has been found" );
}

This is stated very clearly in the online documentation .

If you're looking for a simpler convention when using this approach, you can invert the bits of the operand and convert -1 to 0 and 0 to -1 :

if ( ~$.inArray(14, ["1", "15"]) )

Answer 2

$.inArray returns -1 if item is not found. So jQuery isArray does work properly. That's why when you run your test code:

if ($.inArray(14, testArray)) {
    alert("14 should not be in array. Isn't it?");
}

you see alert. Of course, because -1 is not falsy.