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ajax会返回,但是不刷新当前页面。要强制刷新才管用

WBOY
WBOYOriginal
2016-06-13 11:53:031538browse

ajax能返回,但是不刷新当前页面。要强制刷新才管用

<br /><script type="text/javascript">  <br />	$(document).ready(function() {<br />      $(".quxiao").each(function(){<br />        $(this).click(function(){<br />          quxiao($(this)); <br />        })<br />      })<br />    });<br />    function quxiao(obj){<br />      var oid = $(obj).val();<br />		var oid = $(obj).parent().next().children(".oid").val();//获取节点<br />		//alert(oid);<br />		$.ajax({                        //一个Ajax过程  <br />		type: "post",                   //以post方式与后台沟通  <br />		url : "xgkcg.php",				//与此php页面沟通  <br />		dataType:'json',                //从php返回的值以 json方式 解释  <br />		data: {oid:oid}, <br />		success: function()<br />		{								//如果调用php成功    <br />		alert("成功取消订单!");             <br />		}<br />		});<br />}   <br /></script><br />


php文件
<br />	require_once("config.php");<br />	header('content-type:application/json');<br />	$oid=$_POST['oid'];<br />	$sql="delete from `djs_shops_orders` where oid = '$oid'";<br />	$result = mysql_query($sql);<br />	$row = mysql_fetch_row($result);<br />

html
<br /><td><input id="quxiao" class="quxiao" type="button" value="取消"></td><br />					<td><input type="hidden" class="oid" id="oid" value="<?php echo $v['orderid']; ?>"></td><br />

我强制刷新当前页面他才会显示出来
------解决方案--------------------
1、你的 xgkcg.php 没有输出
2、你执行的是删除指令,mysql_fetch_row($result) 只会报错

------解决方案--------------------
ajax是不能自动刷新页面的,可以如下实现
success: function()<br />        {                              <br />            alert("成功取消订单!");      <br />            window.reload();       <br />        }

或者不刷新的话 调用代码删除
success: function()<br />        {                              <br />            alert("成功取消订单!");      <br />                $(obj).closest('tr').remove();<br />        }

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