关于jquery中的ajaxSubmit问题，请各位大侠帮忙看一下

sun_cathay 2009-06-09 04:34:15

代码如下：
<script type="text/javascript" charset="UTF-8">
$(document).ready(function() {
var options = {
target: '#infotishi', // target element(s) to be updated with server response
beforeSubmit: showRequest, // pre-submit callback
success: showResponse, // post-submit callback
// other available options:
url: '<%=webPath%>/jdmlwhAction.do?method=Add', // override for form's 'action' attribute
type: 'post' , // 'get' or 'post', override for form's 'method' attribute
//data: ,
//dataType: null // 'xml', 'script', or 'json' (expected server response type)
//clearForm: true // clear all form fields after successful submit
//resetForm: true // reset the form after successful submit
// $.ajax options can be used here too, for example:
timeout: 3000
};

// bind to the form's submit event
$('#ajaxsubmit').click(function() {
// inside event callbacks 'this' is the DOM element so we first
// wrap it in a jQuery object and then invoke ajaxSubmit
$(this).ajaxSubmit(options);

// !!! Important !!!
// always return false to prevent standard browser submit and page navigation
return false;
});

$("#Image3").click(function() {
$(this).ajaxSubmit(options);
return false;
});

});//onready

// pre-submit callback
function showRequest(formData, jqForm, options) {
// formData is an array; here we use $.param to convert it to a string to display it
// but the form plugin does this for you automatically when it submits the data
$('#infotishi').show();
$('#infotishi').html('正在保存数据,请稍后......');
$('#infotishi').fadeIn(10000);

var queryString = $.param(formData);
alert(queryString);
var queryString = $('#form :text').fieldSerialize();
alert(queryString);
//var queryString = $('#myForm :text').fieldSerialize();

// jqForm is a jQuery object encapsulating the form element. To access the
// DOM element for the form do this:
// var formElement = jqForm[0];

// here we could return false to prevent the form from being submitted;
// returning anything other than false will allow the form submit to continue
return true;
}

// post-submit callback
function showResponse(responseText, statusText) {
// for normal html responses, the first argument to the success callback
// is the XMLHttpRequest object's responseText property

// if the ajaxSubmit method was passed an Options Object with the dataType
// property set to 'xml' then the first argument to the success callback
// is the XMLHttpRequest object's responseXML property

// if the ajaxSubmit method was passed an Options Object with the dataType
// property set to 'json' then the first argument to the success callback
// is the json data object returned by the server
$(document.forms[0]).ajaxStop(function(){
//异步请求处理停止
$('#infotishi').html('');
$('#infotishi').hide();
});
$(document.forms[0]).ajaxSuccess(function(evt, request, settings){
//异步请求处理成功
$('#infotishi').html('成功');
$('#infotishi').fadeIn(100);
});
$(document.forms[0]).ajaxComplete(function(evt, request, settings){
//异步请求处理完成
$('#infotishi').html('完成');
$('#infotishi').fadeIn(100);
});
$(document.forms[0]).ajaxError(function(evt, request, settings){
//异步请求处理失败
alert("Error");
});

}

</script>

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上面的代码能够进入active里面，但是却获取不到表单的值，并且
var queryString = $.param(formData);
得到的也是空？
请问错在哪里？该如何修改？

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