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我有一个应用程序,其中使用ajax检索了几个div。我想在加载它们时将它们制作成动画,只需将它们从任意位置(顶部,左侧,右侧)滑动到各自的位置即可。我怎样才能做到这一点。这将返回结果的DIV是这个使用jquery动画检索作为ajax响应的div
while($row = mysql_fetch_assoc($agents))
{
$agntId = $row['id'];
$agntDevices = mysql_query("SELECT * FROM devices WHERE agent_id='$agntId'");
$agntProfiles = mysql_query("SELECT * FROM profiles WHERE agent_id='$agntId'");
$noOfDevs = mysql_num_rows($agntDevices);
$noOfPros = mysql_num_rows($agntProfiles);
?>
<div class="agent-tab">
<div class="agent-tab-header"><? echo strtoupper($row['agent_name']); ?></div>
<div class="agent-tab-body">
<? // row starts here ?>
<div class="agent-tab-row">
<div class="agent-tab-side-right">
<div class="agent-tab-side-head agent-content">
No of Profiles
</div>
<div class="agent-tab-side-value agent-content-value">
<?=$noOfPros?>
</div>
</div>
<div class="agent-tab-side-left">
<div class="agent-tab-side-head agent-content">
No of Devices
</div>
<div class="agent-tab-side-value agent-content-value">
<?=$noOfDevs?>
</div>
</div>
</div>
<? // row ends here ?>
<? // row starts here ?>
<div class="agent-tab-row">
<div class="agent-tab-side-right">
<div class="agent-tab-side-head agent-content">
Updated
</div>
<div class="agent-tab-side-value agent-content-value">
<?=$row['updated_on']?>
</div>
</div>
<div class="agent-tab-side-left">
<div class="agent-tab-side-head agent-content">
Added
</div>
<div class="agent-tab-side-value agent-content-value">
<?=$row['added_on']?>
</div>
</div>
</div>
<? // row ends here ?>
</div>
</div>
<? } ?>
使用检索这个和动画jQuery的IM如下。(目前它没有动画)
function LoadDashBoard()
{
var loadUrl = "ajax/load_agents.php";
var ajax_load_bar = "<div align='center'><img src='images/loadingAnimation.gif' alt='loading...' /></div>";
$("#actdiv").html(ajax_load_bar).load(loadUrl);
$(".agent-tab").animate({
top: "+=250px",
}, 1000);
}
谢谢...这工作我不知道该属性必须隐藏,以便它可以轻松地动画 – swordfish 2011-05-27 15:57:05