Having trouble with Jquery's .mouseover() method? Let's catch the solution!

I'm currently working on a jQuery homework assignment and I've been attempting to enhance the mouseover feature, but unfortunately, it's not functioning as expected.

$("button").mouseover(function() {
  $(this).css({
    left: (Math.random() * 300) + "px",
    top: (Math.random() * 300) + "px",
  }, 1);
});
button {
  position: absolute;
  top: 400px;
  left: 500px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<body>
  <h1>Catch me if you can !</h1> 
  <div class="main"></div>
  <button>
    <img src="cat/grey-cat-001.gif">
  </button>
</body>

Answer №1

The reason for this behavior is that the default position setting for elements is "static," which does not respond to styling with top, bottom, left, or right properties.

By default, HTML elements are statically positioned.

Elements with static positioning do not react to changes in the top, bottom, left, and right properties.

To fix this issue, you simply need to include either "position: relative" or "position: absolute" for the button element, as demonstrated in the following code snippet:

$("button").mouseover(function() {
  $(this).css({
    left: (Math.random() * 300) + "px",
    top: (Math.random() * 300) + "px",
  }, 1);
});
button {
  position: relative;  
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<body>
  <div class="main"></div>
  <button>
    <img src="cat/grey-cat-001.gif">
  </button>
</body>

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