weird output of the combination of $.each and jQuery.inArray function

Question

My complete code:

jQuery.extend({

combinationCheck: function (p1position) {

    var Combination = [1, 2, 3, 4, 5, 6, 7, 8];
    Combination[0] = [1, 2, 3];
    Combination[1] = [4, 5, 6];
    Combination[2] = [7, 8, 9];
    Combination[3] = [1, 4, 7];
    Combination[4] = [2, 5, 8];
    Combination[5] = [4, 6, 8];
    Combination[6] = [1, 5, 9];
    Combination[7] = [3, 5, 7];


    $.each(p1position, function (index, value) {

        var num = value;

        if ($.inArray(String(value), Combination[1]) != '-1') {
            alert("there");
        }
        else {
            alert("not there");
        }

    });
});

so it works. If I were to set num to 5, it alerts "is there", and for 8 --> "not there". but the problem is I have another array.

p1position = [1,5];

and go through the array..

$.each(p1position,function(index,value){
    var num = value;
//then call the jQuery.inArray function as written above, it always return not there. even though 5 is in the Combination[1] array.
});

I am so confused of trying to solve this problem for hours.

Why in heaven's name are you comparing the result of `inArray` to a **string**?! — T.J. Crowder, Jan 31 '12 at 09:27
Also, why initialize the array with values you're just going to overwrite immediately afterward? — T.J. Crowder, Jan 31 '12 at 09:28
because I am programming a TicTacToe (vs PC). there 1st players postion will be saved in an array. and the elements in this array will be checked with Combination. then I take combination, program it, so the PC responds and plays logically. — kk-dev11, Jan 31 '12 at 09:52
can you please share your complete code. It worked for me . see http://jsfiddle.net/gApqm/4/ — Diode, Jan 31 '12 at 10:05
@Freetalk13: I'm afraid you haven't understood either of my comments. 1. The result of `inArray` is a number; compare it against a number, not a string. 2. Doing `a = [1, 2, 3, 4];` when you're going to immediately follow it with `a[0] = ...; a[1] = ...; a[2] = ...; a[3] = ...;` is pointless, you're overwriting the data you used to initialize it. Just use `a = [];` to start with. Or better yet, initialize it all at once: http://pastie.org/3288165 — T.J. Crowder, Jan 31 '12 at 11:16
@T.J.Crowder how can I check a number is there? I tried !=-1 .. It doesn't even work. — kk-dev11, Jan 31 '12 at 11:28

T.J. Crowder · Answer 1 · 2012-01-31T11:52:30.663

The specific problem with your code that you're asking about is that you're turning the value into a string before the check:

if ($.inArray(String(value), Combination[1]) != '-1') {
//            ^^^^^^^^^^^^^

inArray does an === (strict equality) check, and "1" !== 1. That line should read:

if ($.inArray(value, Combination[1]) !== -1) {

Changes:

Don't turn the value into a string.
Compare the result with -1 (a number), not "-1" (a string). inArray returns a number.
Use !== rather than != (this is mostly a matter of style, you can use != if you prefer).

There are several other problems with that code, though.

You're missing a }, so the code you say is your complete code doesn't parse.
You're re-creating Combination every time the combinationCheck is called. If your goal is to create a Tic-Tac-Toe game, you're going to need to be able to retain the Combination state between checks.

Here's a fairly minimal set of fixes:

(function() {
    var Combination = [
            [1, 2, 3],
            [4, 5, 6],
            [7, 8, 9],
            [1, 4, 7],
            [2, 5, 8],
            [4, 6, 8],
            [1, 5, 9],
            [3, 5, 7]
        ];

    jQuery.extend({

        combinationCheck: function (p1position) {

            $.each(p1position, function (index, value) {

                if ($.inArray(value, Combination[1]) !== -1) {
                    alert(value + " is there");
                }
                else {
                    alert(value + " is NOT there");
                }

            });
        }
    });

})();

...which given:

jQuery.combinationCheck([1, 5]);

...reports that 1 is not found, but 5 is.

Live copy

hmm.. i tried, but still doesn^t work for me. I have the exactly the same code as you've written — kk-dev11, Jan 31 '12 at 12:37
@Freetalk13: *"I have the exactly the same code as you've written"* Apparently not, as it does work (see the live link). The only thing I can think of is that there's something *else*, unrelated to the quoted code, that's preventing it from running at all. Look in the JavaScript console for errors. — T.J. Crowder, Jan 31 '12 at 13:58

weird output of the combination of $.each and jQuery.inArray function

1 Answers1