Simple jQuery Hover Menu

Question

This looks really bloated, can it be rewritten any better/more compact:

$("#cart-summary").hover(
      function () {
        $('.flycart').slideDown('fast');
      }
   );
   $(".flycart").hover(
      function () {}, // mousein function not required
      function () { // hide menu on mouseout
        $('.flycart').slideUp('fast');
      }
   );
    $('.flycart a.close').click(function(){
        $(this).parents('.flycart').hide();
    });

Thanks!

Nope, compared to the accepted answer. :-)

eozzy
– eozzy

2009-10-01 14:53:56 +00:00
Commented Oct 1, 2009 at 14:53 — eozzy
– eozzy, Commented Oct 1, 2009 at 14:53

Alex Gyoshev · Accepted Answer · 2009-10-01 15:59:56Z

2

$("#cart-summary").mouseenter(function () {
    $('.flycart').slideDown('fast');
});

$(".flycart")
    .mouseleave(function () {
        $(this).slideUp('fast');
    })
    .find('a.close')
        .click(function(){
            $(this).parents('.flycart').hide();
        });

It's a small improvement, though. I couldn't guess the relationship between the #cart-summary and the .flycarts.

edited Oct 1, 2009 at 15:59

answered Oct 1, 2009 at 14:18

Alex Gyoshev

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Comments

Eric · Accepted Answer · 2009-10-01 14:21:57Z

1

In short, no. However, you can do without the empty hover function: just use mouseenter() and mouseleave(). mouseover and mouseout have subtle differences to mouseenter and mouseleave. Look at the jQuery API for more info.

$("#cart-summary").mouseenter(function()
{
    $('.flycart').slideDown('fast');
});

$(".flycart").mouseleave(function()
{
    $(this).slideUp('fast');
});

$('.flycart a.close').click(function()
{
    $(this).parents('.flycart').hide();
});

answered Oct 1, 2009 at 14:21

Eric

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