javascript image rollover inside div
Question
i have a single div 100px X 300px. What's the easiest way in JavaScript so when I hover over the div i show an image and then when i leave the div the image disappears.
for starters i thought the following would get me started but i can't seem to remove the image properly
<script type="text/javascript">
function MouseOver_Event(elementId) {
var imgToCreate = document.createElement('img');
imgToCreate.setAttribute('id', 'imgHandle');
imgToCreate.setAttribute('src', elementId + '.png');
imgToCreate.setAttribute('onmouseout', 'MouseOut_Event('+elementId+')');
var targetDiv = document.getElementById(elementId);
targetDiv.appendChild(imgToCreate);
targetDiv.removeAttribute('onmouseover', 'MouseOver_Event');
}
function MouseOut_Event(elementId) {
var imgToRemove = document.getElementById('imgHandle');
var targetDiv = imgToRemove.parentNode();
if (imgToRemove != null)
targetDiv.removeChild(imgToRemove);
targetDiv.setAttribute('onmouseover', 'MouseOut_Event(' + elementId + ')');
}
</script>
</head>
<body>
<div id="div1" onmouseover="MouseOver_Event(this.id)"></div>
<div id="div2" onmouseover="MouseOver_Event(this.id)"></div>
<div id="div3" onmouseover="MouseOver_Event(this.id)"><img src="Div3.png" alt="test" onmouseout="MouseOut_Event(parentNode's id or something)" /></div>
</body>
2 answers
solution1
2 2011-10-06 03:09:19
You're attaching your MouseOut_Event to onmouseover instead of onmouseout. But you probably don't need to be messing with dynamic event creation anyway; just add onmouseout="MouseOut_Event(this.id)" to the three divs and that should do it.
solution2
1 ACCPTED 2011-10-06 03:38:16
Why don't you use CSS instead ?
For example:
#div1{background:none;}
#div1:hover{background:url('src/div1.png') no-repeat;}
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