1.onsubmit
form表单的onsubmit方法在submit执行之前提交表单
<script type="text/javascript"> ?????function sub() { ?????????// jquery 表单提交 ??????????$("#formId").ajaxSubmit(function(message) { ??????????// 对于表单提交成功后处理，message为返回内容 ??????????}); ??????????return false; // 必须返回false，否则表单会自己再做一次提交操作，并且页面跳转 ??????} ??</script> ?<form id="formId" action="servlet" method="post"onsubmit="return sub();"> ??????<input type="submit" value="提交"/> ??</form>2.ajax提交
<script type="text/javascript"> ?????function sub() { ?????????$.ajax({ ?????????????????cache: true, ?????????????????type: "POST", ?????????????????url:"Servlet", ?????????????????data:$(‘#formId‘).serialize(),// 你的formid ?????????????????async: false, ?????????????????error: function(request) { ?????????????????????alert("Connection error:"+request.error); ?????????????????}, ?????????????????success: function(data) { ?????????????????????alert("SUCCESS!"); ?????????????????} ?????????????}); ?????} ?</script> ?<form id="formId" ?method="post""> ?????<input id="input1" /> ??????<input id="input2" /> ??????<input id="input3" /> ?????<input type="button" value="提 交" onclick="sub()"/> ?</form>注意问题：
button 如果设置为 type="submit"，会出现提交两次的问题，请把submit设置为type="button"
ajax提交表单数据不跳转
原文地址：https://www.cnblogs.com/woniu666/p/9901776.html