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Using jQuery delay() with css()

Asked 2011-03-22T18:34:18.500
57

Why does delay() work here:

$('#tipper').mouseout(function() {
  $('#tip').delay(800).fadeOut(100);
});

But this fails to delay:

$('#tipper').mouseout(function() {
  $('#tip').delay(800).css('display','none');
});

// EDIT - here's a working solution

// EDIT 2 - some bugs fixed

$('#tipper').mouseleave(function() {
  setTimeout( function(){
    $('#tip').css({'display','none'});
       },800);
});
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2 Answers

138

delay() works with the animation (fx) queue. Changing a css property does not work via that mechanism, and thus is not affected by the delay directive.

There is a workaround -- you can inject the property change as a queued operation, like this:

$('#tip')
  .delay(800)
  .queue(function (next) { 
    $(this).css('display', 'none'); 
    next(); 
  });

Also, you should probably be using .hide() instead of .css('display','none').

Here's a working example: http://jsfiddle.net/redler/DgL3m/

answered 2011-03-22T18:36:44.460
12

Added to Ken Redler's answer / suggestion:

Also, you should probably be using .hide() instead of .css('display','none').

You can do :

$('#tip').delay(800).hide(0);

The 0 is important here. Passing a value to .hide() will implicitly add it to the fx queue and therefore, this will work like expected.

answered 2011-03-22T18:44:25.997

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