我在这里有一个拖放代码来自jquery我想获取图像的值然后将其插入数据库,当图像从可放置区域中删除然后更新并删除图像值..我怎么能做到了吗? 这是代码......
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>jQuery UI Droppable - Default functionality</title>
<link rel="stylesheet" href="../../themes/base/jquery.ui.all.css">
<script src="js/jquery-1.4.4.js"></script>
<script src="js/jquery.ui.core.js"></script>
<script src="js/jquery.ui.widget.js"></script>
<script src="js/jquery.ui.mouse.js"></script>
<script src="js/jquery.ui.draggable.js"></script>
<script src="js/jquery.ui.droppable.js"></script>
<link rel="stylesheet" href="js/demos.css">
<style>
#comment { width: 100px; height: 100px; padding: 0.5em; float: left; margin: 10px 10px 10px 0; }
#draggable1 { width: 100px; height: 100px; padding: 0.5em; float: left; margin: 10px 10px 10px 0; }
#droppable { width: 500px; height: 200px; padding: 0.5em; float: left; margin: 10px; background: silver;}
</style>
<script>
$(function() {
$( "#comment" ).draggable();
$( "#draggable1" ).draggable();
$( "#droppable" ).droppable({
drop: function( event, ui ) {
$( this )
.find( "p" )
.html("Dropped!")
},
out: function(event, ui) {
$(this)
.find( "p" )
.html("Drop Node Here!");
}
});
});
</script>
</head>
<body>
<div class="demo">
<div id="comment" class="ui-widget-content">
<img src="images/signup.png" id="1">
</div>
<div id="draggable1" class="ui-widget-content">
<img src="images/signup.png" id="2">
</div>
<div id="droppable" class="ui-widget-header">
<p>Drop Node here</p>
<?php
$comment = "#comment";
$drop = "#droppable";
$dropped = "dropped";
require_once("includes/connection.php");
require_once("includes/close.php");
?>
</div>
</div><!-- End demo -->
</body>
</html>
答案 0 :(得分:0)
在我的脑海中,这样的东西会给你图像的src
drop: function( event, ui ) {
var image_src = $(ui).attr('src');