Hi there ,
I am trying to make Images on my site auto-ZoomIn on "mouseover" event , and auto-ZoomOut on "mouseout" event , but this is not working properly.
I hope some of you guys have tried such thing or seen such thing espescially on MAC , where Icons performs the same behaviour on mouseover, but i want to do it with Images .
It looks pretty simple , but becomes tough to combine Javascript with my PHP Code , where the Image Source is taken from the MySql Database Query.
Major Problems:
1-How to transfer Image Source from a PHP " $ " Variable to the Javascript Variable " var " ?
2- Should I make 6 <div>'s for 6 Images, in order to show their Zoomed HQ Image?
3-I don't want to use jQuery or other plugin files to make this happen for now.
Here is my Code :
Code:
<html>
<head>
<style>
#productImg
{
width:600px;
height:auto;
position:relative;
}
#productImgBeta
{
width:auto;
height:auto;
position:absolute;
z-index:-100000;
visibility:hidden;
}
</style>
<script language="javascript" type="text/javascript">
function imageZoomCancel()
{
document.getElementById("productImgBeta").style.zIndex = "-100000";
document.getElementById("productImgBeta").style.visibility = "hidden";
}
function imageZoom()
{
document.getElementById("productImgBeta").style.zIndex = "100000";
document.getElementById("productImgBeta").style.visibility = "visible";
}
function imgSrc()
{
//Not Sure what to put in here
}
</script>
</head>
<body>
<div id="productImg">
<div id="productImgBeta">
<?php
// $row['ProductImg'](0-11) are taken from MySQL Database
// $row['ProductImg'](0-11) Contain the relative paths of the Images
$LQImage1= $row['ProductImg']; $HQImage1= $row['ProductImg6'];
$LQImage2= $row['ProductImg1']; $HQImage1= $row['ProductImg7'];
$LQImage3= $row['ProductImg2']; $HQImage1= $row['ProductImg8'];
$LQImage4= $row['ProductImg3']; $HQImage1= $row['ProductImg9'];
$LQImage5= $row['ProductImg4']; $HQImage1= $row['ProductImg10'];
$LQImage6= $row['ProductImg5']; $HQImage1= $row['ProductImg11'];
?>
//Here the Zoomed In(High Quality) Image will be displayed
<img align="left" width="500px" height="400px" src="imgSrc()" />
</div>
//Here The Zoomed Out(Low Quality) Image will be displayed
<img onmouseover="imageZoom()" onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage1); ?>" />
<img onmouseover="imageZoom() onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage2); ?>" />
<img onmouseover="imageZoom()" onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage3); ?>" />
<img onmouseover="imageZoom()" onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage4); ?>" />
<img onmouseover="imageZoom()" onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage5); ?>" />
<img onmouseover="imageZoom()" onmouseout="imageZoomCancel()" width="150px" height="150px" src="<?php echo ($LQImage6); ?>" />
</div>
</body>
</html>
I hope you understand the Code , if you find any Error regarding this Code or you think this Code needs improvements , then please guide me with my mistakes.
Please try this Code in your Browsers , i have tried it in IE8, Chrome 14(beta), Opera 11 and Firefox 6 , but all gave the same result.
Note : This Code is extracted from my site, it is not a complete page of my site but a small part of it , so it may have issues in the display.