jQuery mouseenter() and mouseleave() functions work repeatedly

Baris Demirel picture Baris Demirel · May 17, 2014 · Viewed 30.3k times · Source

I have a picture and a div. The div is hidden(display:none;). I want to show the div when mouse is over the picture and hide the div again when mouse is not over the picture. I use mouseenter() and mouseleave() events to do this but when moue is over the picture, both of the functions work repeatedly. Here is the code that I define functions:`

$("#pic").mouseenter(function(){
  $("#checkin").slideDown();
});
$("#pic").mouseleave(function(){
  $("#checkin").slideUp();
});

I also tried the hover method but the result is the same.

$("#pic").hover(
  function(){
    $("#checkin").slideDown(200);
  },
  function(){
    $("#checkin").slideUp(200);
  }
);

What is the problem, I can't understand.

Update: Here is the HTML code

<tr><td valign='top' class='checkinpic'><img src='img.png' id='pic' height='100%'></td></tr>

...

<div id='checkin'>
You are not going to any activity this week.
</div>

Answer

Baris Demirel picture Baris Demirel · May 17, 2014

I found the solution. When I changed the element that works with the mouseleave event, it worked:

var block = false;
$("#pic").mouseenter(function(){
if(!block) {
    block = true;
    $("#toggleDiv").slideDown(400, function(){
        block = false;
    });
}
});
$("#pic").mouseleave(function(){
if(!block) {
    block = true;
    $("#toggleDiv").slideUp(400, function(){
        block = false;
    });
}
});

Thank you for your help.

Update: I noticed that giving both the picture and div a class and defining the mouseenter and mouseleave events of the class is a better solution.