Error with jQuery syntax?

Question

Based on an answer from the stack overflow post at: Change background color on mouseover and remove it after mouseout

the below code should set the css to display: block; on hover and display: none; when they hover off.

Can anyone see anything wrong with the below code? (No console errors are occuring) I'm basically trying to make myself a simple tool tip.

$(function () {
    $(document).on('hover', '.inter [class]', function () {
        $('._22t').css({
            'display': 'block'
        });
    }, function () {
        $('._22t').css({
            'display': 'none'
        });
    });
});

`on()` only takes one function argument, not two. Read the documentation. — Barmar, May 24 '13 at 03:05
maybe you should try `$('._22t').toggleClass('your_css_class');` — Þaw, May 24 '13 at 03:07
So simply changing it to $('.inter [class]').hover() would fix it? — sephiith, May 24 '13 at 03:08
That will work if all the elements exist at the time you run it. — Barmar, May 24 '13 at 03:09

score 7 · Accepted Answer · answered May 24 '13 at 03:07

on doesn't accept 2 callback functions and apart from that you can't use hover pseudo event name with the on method:

Deprecated in jQuery 1.8, removed in 1.9: The name "hover" used as a shorthand for the string "mouseenter mouseleave". It attaches a single event handler for those two events, and the handler must examine event.type to determine whether the event is mouseenter or mouseleave. Do not confuse the "hover" pseudo-event-name with the .hover() method, which accepts one or two functions.

$(document).on('mouseenter mouseleave', '.inter [class]', function(event) {
    $('._22t').toggle(event.type === 'mouseenter');
});

Error with jQuery syntax?

1 Answers1