ajax提交整个from表单示例代码

$.ajax( {
type : "POST",
url : "<%=request.getContextPath()%>/control/daControlAction_updateEmotecontrol.action",
data : $("#form1").serialize(),
dataType : "text",
success : function(data) {

if (data > 0) {

ymPrompt.alert('修改成功！', null, 200, '提示!',
closeWindow);
window.opener.location.reload();
}
},
beforeSend : function() {

},
complete : function() {

}
});