CODEWITHSUNDEEP
jqueryjquery-callback
Bram Vanroy
Bram Vanroy

Reputation: 28475

Understanding stop() and callbacks in jQuery

I have this fiddle with a function like this:

$("#success").stop(true, true).css({
    "top": offT,
        "opacity": 0
}).animate({
    "top": "-=40",
        "opacity": 1
}, 700, function () {
    $(this).delay(4000).animate({
        "opacity": 0
    }, 1200);
});

When you click a button once, there's no problem. But try clicking a button and after roughly two seconds click another. You'll notice that rather than delaying the fade out for 4 seconds, the div hides much sooner. In other words, the callback initiated by the first click is still active when clicking on the second button. I find this strange, because I call stop(true,true). Doesn't this prevent the callback from the first click to be executed and finished? If not, how can I force it to do so?

Upvotes: 1

Views: 136

Answers (1)

Genhis
Genhis

Reputation: 1524

It should prevent the callback to be executed, but after your second click the callback has already been executed - the delay is in queue.

To prevent this you can use JS window.setTimeout() method and cancel the delay.

See this answer: How to stop the delay.

var delay;
$("button").click(function () {
    var offT = $(this).offset().top;
    window.clearTimeout(delay);
    $("#success").stop(true, false).css({
        "top": offT,
            "opacity": 0
    }).animate({
        "top": "-=40",
            "opacity": 1
    }, 700, function () {
        delay = window.setTimeout(function() {
            $("#success").animate({
                "opacity": 0
            }, 1200);
        }, 4000);
    });
});

Upvotes: 2

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