我是学习jquery.ajax的新手.我想将一些数据从a.php粘贴到b.php.这是我的代码:
a.php只会
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js" type="text/javascript"></script>
<script language="javascript">
$(document).ready(function () {
var params = "value=" + $('#send').text();
$.ajax({
url:'b.php',
type:'post',
dataType:'html',
data:params,
success:function(data){ $("#result").html($(data).html('#aa')); }
});
});
</script>
<body>
<div id="result"></div>//sometimes the return data is empty, so this part just return: <div="aa"></div>
<div id="send">apple</div>
b.php
<?php
echo '<div id="aa">';
//'.$_REQUEST['value'].' will put in some process here, but sometimes the return data is empty.
echo '</div>';
?>
如何以及在何处添加判断,如果<div id="aa"></div>为空,则sorry, there is no result在它之间添加
所以a.php,信息将显示:<div id="result"><div id="aa">sorry, there is no result</div></div>
谢谢.
首先,您应该始终返回DATA/INFORMATION,而不是HTML标签或代码......您在HTML部分中执行此操作.
<?php
//send ONLY the data, best to send in JSON syntax
?>
然后你用
$.ajax({
url:'b.php',
type:'post',
dataType:'html',
data:params,
success: function(data) {
if(data.length == 0)
$("#result").html("<span class='no-data'>No Data</span>");
else
{
// Loop through the data and add it as, for example an <li> in a <ul>
}
}
});
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