step1 batch:(下拉)在动态加载,选择一个值从下拉步骤2是通过ajax调用加载后,在set2当我点击编辑按钮,步骤3是通过ajax调用再次加载。在步骤3中,当我点击编辑按钮ajax调用工作正常,但它不是张贴值到PHP脚本。
//ajax call
function validateFees(strAddNo) {
var collectFees = $("#collectFees").val();
if(collectFees == "")
{
$("#validateFeesResult").html('<div class="info">Please enter your Fees Amount.</div>');
$("#collectFees").focus();
}
else
{
var dataString = 'collectFees' + collectFees + 'strAddNo' + strAddNo;
$.ajax({
type: "POST",
url: "validateFees_script.php",
data: dataString,
cache: false,
beforeSend: function()
{
$("#validateFeesResult").html('Loading...');
},
success: function(response)
{
$("#validateFeesResult").hide().fadeIn('slow').html(response);
}
});
}
}
我相信这是非常简单的,但我不明白如何做到这一点?
将数据格式更改为:
var dataString = 'collectFees=' + collectFees + '&strAddNo=' + strAddNo;
查看:http://api.jquery.com/serialize/