通过ajax POST传值，但返回为空解决思路-PHP Tutorial-php.cn

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请选择班级：

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通过ajax POST传值，但返回为空解决思路

Jun 13, 2016 am 11:54 AM

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通过ajax POST传值，但返回为空
第一次接触ajax，想通过一个简单的例子来练练
index.php中有一个下拉列表，选择不同的班级返回不同的学生信息
代码如下：

<?php<br />	//从数据库把班级名称取出来放入下拉列表中<br />	mysql_connect("localhost","root","root") or die("数据库连接失败");<br />	mysql_select_db("studentmanage") or die("数据库不存在");<br />						<br />	$sql="select * from class";<br />	$rs=mysql_query($sql);<br />	$info=array();<br />	while($row=mysql_fetch_assoc($rs)){<br />		$info[]=$row;<br />	}<br /><br />?><br /><!doctype html><br /><html><br />	<head><br />		<title>查询学生信息</title><br />		<meta http-equiv="content-type" content="text/html;charset=utf-8"/><br />		<style type="text/css"><br />		<br />		</style><br />		<script type="text/javascript"><br />			var xmlhttprequest;;<br />			function getMessage(){<br />				<br />				if(window.ActiveXObject){<br />					xmlhttprequest=new ActiveXObject("Microsoft.XMLHTTP");<br />				}else{<br />					xmlhttprequest=new XMLHttpRequest();<br />				}<br />				<br />				if(xmlhttprequest){<br />					var url="result.php";<br />					var cid=document.getElementById("class").value;<br />					var show=document.getElementById("info");<br />					var data="class=cid";<br />					xmlhttprequest.open("post",url,true);<br />					xmlhttprequest.setRequestHeader("content-type","application/x-www-form-urlencode");<br />					xmlhttprequest.onreadystatechange=function(){<br />						if(xmlhttprequest.readyState==4){<br />							if(xmlhttprequest.status==200){<br />								var info=xmlhttprequest.responseText;<br />								if(info==""){<br />									show.value="你好，你查询的班级暂时还没有学生";<br />								}else{<br />									show.value=info;<br />								}<br />							}<br />						}else{<br />							alert("error");<br />						}<br />					}<br />					xmlhttprequest.send(data);<br />				}<br />			}<br />		</script><br />	</head><br />	<body><br />		<div id="content"><br />			<h1 id="br-请选择班级-br-select-nbsp-name-class-nbsp-id-class-nbsp-onchange-getMessage-br-option-nbsp-value-nbsp-selected-choose-nbsp-class-option-br-php-nbsp-foreach-info-nbsp-as-nbsp-v-br-option-nbsp-value-php-nbsp-echo-nbsp-v-cid-php-nbsp-echo-nbsp-v-cname-option-br-php-nbsp-br-select-br"><br />				请选择班级：<br />				<select name="class" id="class" onchange="getMessage()"><br />					<option value="" selected>choose class</option><br />					<?php foreach($info as $v){?><br />					<option value="<?php echo $v['cid'];?>"><?php echo $v['cname'];?></option><br />					<?php }?><br />				</select><br />			</h1><br />			<div id="info"></div><br />		</div><br />	</body><br /></html>

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通过result.php查询数据库，并返回学生信息，代码如下：

<?php<br />	//header("content-type:text/html;charset=utf-8");<br />	header("Content-type:text/xml;charset=utf-8");<br />	header("Cache-Control:no-cache");<br />	$class=$_POST['class'];<br />	<br />	mysql_connect("localhost","root","root") or die("数据库连接失败");<br />	mysql_select_db("studentmanage");<br />	mysql_query("set names utf-8");<br />	$sql="select * from student where cid='$class'";<br />	//$sql="select * from student where cid=3";<br />	$rs=mysql_query($sql);<br />	$info=array();<br />	while($row=mysql_fetch_assoc($rs)){<br />		$info[]=$row;<br />	}<br />	<br />	$result="<table><tr><td>学号</td><td>姓名</td><td>住址</td><td>班级</td></tr>";<br />	foreach($info as $v){<br />		$result.="<tr><td>{$v['sid']}</td><td>{$v['name']}</td><td>{$v['address']}</td><td>{$v['cid']}</td></tr>";<br />	}<br />	$result.="</table>";<br />	echo $result;<br />?>

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我单独测试了下result.php，我确定没有问题，能从数据库中取出对应的信息，但是就是不能把取到的信息返回到index.php中
求各位帮帮忙，找找问题在哪

------解决方案--------------------
给div设置value?即使你的查询时正常的，由于div的value并不会直接在视图中展现，所以你看不到效果。
你需要的应该是innerHTML

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