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() { 

} 
});