jqueryForm的ajaxsublime方法

xiaoxiao2021-02-28  107

先引入jqueryform

<script type="text/javascript" src="__PUBLIC__/js/jquery.form.js"></script>

html

<form class="myform " onsubmit="return false;"> <input type="text" name="id" value="{$data.id}"> <input class="btn btn-white" type="submit" id="sub" value="确定"> </form>

js部分

$("#sub").on('click',function () { $(".myform").ajaxSubmit({ type: "post", dataType: "json", success: function (data) { if (data.status === 1) { dialog({content: data.info}).showModal(); setTimeout(function () { window.location.href="{:U('Order/repair')}"; }, 300); } else { dialog({title: '友情提示', content: data.info, ok: true}).showModal(); } } }); })
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