Jquery Datepicker, at wits end

Question

This code used to work but now it doesn't and it's only taking the first value in the array...

var unavailableDates is an array that stops dates showing on the datepicker..

any ideas??

It's not cycling through all dates in the array for some reason!?!?

var unavailableDates = ["4-7-2011","5-7-2011"];

function unavailable(date) {
    var dmy = date.getDate() + "-" + (date.getMonth() + 1) + "-" + date.getFullYear();
    if ($.inArray(dmy, unavailableDates) == 0) {
        return [false, "", "Unavailable"];
    } else {
        var day = date.getDay();
        return [(day != 0 && day != 2 && day != 3 && day != 4 && day != 6)];
    }
}

$(function(){

    $('#smh').datepicker({
        showOn: "both",
        buttonImage: "images/calendar.gif",
        buttonImageOnly: true,
        beforeShowDay: unavailable,
        minDate: -0,
        dateFormat: "dd/mm/yy",
        onSelect: function(e) {
        e = e.split('/')[1] + '/' + e.split('/')[0] + '/' + e.split('/')[2];
        var date = new Date(e);
        var day = date.getDay(); // 0 = sunday etc...        
        if (day === 1) {
            $("#check2").hide();
            $("#text").hide();
            $("#check1").show();

        } else if (day === 5) {
            $("#check1").hide();
            $("#text").hide();
            $("#check2").show();

        } 
        $("#bdate").html(this.value);
    } 
    })

Answer 1

What happens is $.inArray returns the index of the found item(if it finds it, else -1)

so youll need to check if the index is not -1

...$.inArray(dmy, unavailableDates) != -1...

Answer 2

In jQuery the $.inArray(elem, array) method returns -1, and not 0, when elem is not found in array.

So I think you should use:

if ($.inArray(dmy , unavailableDates) == -1)

instead of:

if ($.inArray(dmy, unavailableDates) == 0)