jQuery inArray is always returning -1

Question

I cannot figure out why I keep getting -1 for lastProductIndex when clearly the lastProductID is in the array!

var lastProductID = 6758;
var allProductIDs = [5410, 8362, 6638, 6758, 7795, 5775, 1004, 1008, 1013]
var lastProductIndex = $.inArray(lastProductID, allProductIDs);

Answer 1

What does allProductIDs.indexOf(lastProductID) return?

Answer 2

Are you using any other JavaScript libraries in your page?

There could be a conflict for the $ shorthand if you are. This can be solved in a number of ways, one of which would be wrapping the code in a self-executing anonymous function

(function($) {
    var lastProductIndex = $.inArray(lastProductID, allProductIDs);
})(jQuery);

Answer 3

Here is why everyone hates languages that are not TYPED. I had initially set the lastProductIndex value with a value, but it was a string (because I had gotten the value from an HttpResponse object from returned JSON. So consequently I had set the variable to a string because of the returned JSON value was a string. When I hard coded the number 6758 into $.inArray it worked fine so that caught my attention.

Answer 4

This confused me too at first, I had an identical problem. It seems jQuery.inArray() requires a string. So you could just do:

var lastProductID = "6758";

In my case I was trying to do:

var foo = date.getTime()/1000;

Which always resulted in -1 being returned from $.inArray()

But you can just cast it to a string:

var foo = String(date.getTime()/1000);

Hope that helps somebody :)

Answer 5

Try this instead:

$.grep(allProductIDs, function(n) { return n == lastProductID; });

Caveat: grep returns an array.

It looks like jQuery does an === instead of == with inArray.

Answer 6

I had the same problem yesterday,,

var data=[2,4,6,8];
alert( $.inArray(4,data) ) // output 1 as expected
alert( $.inArray("4",data) ) // output -1 as expected

Generally this occurs when you get the value to check from a input element or something that returns a string. You need to do parseInt() to convert it to a number...

Answer 7

Is it the missing semicolon at the end of the 2nd line?