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Home/ Questions/Q 4077248
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Editorial Team
  • 0
Editorial Team
Asked: May 20, 20262026-05-20T17:31:36+00:00 2026-05-20T17:31:36+00:00

I’m building a form to make a reservation on a website. When you choose

  • 0

I’m building a form to make a reservation on a website. When you choose a shift, the available dates are loaded. Now I’d like to load the available seats when you pick a date.

//Shift choosen
$("input[name$=shift]").change(function() {
    $.ajax({
        type: 'POST',
        url: '/reservations/date/list/future-notfull-withshift',
        data: $("input[name$='shift']").serialize(),
        success: function(msg) {
                $("#date").html(msg);
                $("#date").fadeIn(250);
            }
    });
});

//Date choosen
$("input[name$=dates]").live("change", function() {
    $.ajax({
        type: 'POST',
        url: '/reservations/date/list/availableseats',
        data: $("input[name$='dates']").serialize(),
        success: function(msg) {
            $("#seats").html(msg);
            $("#seats").fadeIn(250);
        }
    });
});

In this code the first block works just fine, but the second call, which is very much the same, doesn’t work: No Ajax call is made …
Any idea why? How can I solve this?

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  1. Editorial Team
    Editorial Team
    2026-05-20T17:31:37+00:00Added an answer on May 20, 2026 at 5:31 pm

    Be sure that $("input[name$=dates]") find something. Maybe jQuery can’t find your dates input?

    BTW: is it your application? If you can – use id attrib or class attrib. input[name$=dates] selector is not very fast. Even scoped find is better: $('#foo').find("input[name$=dates]").live(...).
    Also take a look at delegate function in jQuery. Its like live but better 😉

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