php结合javascript弹出图片浏览

php查询数据库显示小图，点击小图弹出大图

<?php 
//Get all the images in the database (use a WHERE clause to limit what it finds) 
$sql = mysql_query("SELECT * FROM database_name"); 
//Start A counter we will start at 1 
$i = 1; 
//Get all the returned results, save them in an array, and loop throught the array 
//to display all the returned results 
while($row = mysql_fetch_array($sql)){ 
    //Create a list from the returned image file, and make a $width and $height variable 
    list($width, $height) = getimagesize("url/to/large/image/".$row['img_file_name']); 
    //Add 30 to the returned width and height so it will fit in the popup window 
    //If it still doesn't fit, change the two thirtys to a larger number. 
    $width += 30; 
    $height += 30; 
    //Use javascript to make the popup 
    //Everytime the script runs through, $i will change so we can have dynamic popup window sizes 
    echo " 
    <script type="text/javascript"> 
        function load".$i."(){ 
            var load = window.open('url/to/large/image/".$row['img_file_name']."', ' ', 'scrollbars=no,menubar=no,height=".$height.",width=".$width.",resizable=no,toolbar=no,location=no,status=no'); 
        } 
    </script>"; 
    //This will display our thumbnail on the screen and make a link to the popup function 
    //that was just created in the above javascript. 
    echo' 
    <p> 
        <a href="javascript:load'.$i.'()"> 
        <img src="url/to/thumb/image/'.$row['img_file_name'].'" /> 
        </a> 
    </p>'; 
    //Add one to $i 
    $i++; 
} 
?>