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 //从数据库把班级名称取出来放入下拉列表中 mysql_connect("localhost","root","root") or die("数据库连接失败"); mysql_select_db("studentmanage") or die("数据库不存在"); $sql="select * from class"; $rs=mysql_query($sql); $info=array(); while($row=mysql_fetch_assoc($rs)){ $info[]=$row; }?><!doctype html><html> <head> <title>查询学生信息</title> <meta http-equiv="content-type" content="text/html;charset=utf-8"/> <style type="text/css"> </style> <script type="text/javascript"> var xmlhttprequest;; function getMessage(){ if(window.ActiveXObject){ xmlhttprequest=new ActiveXObject("Microsoft.XMLHTTP"); }else{ xmlhttprequest=new XMLHttpRequest(); } if(xmlhttprequest){ var url="result.php"; var cid=document.getElementById("class").value; var show=document.getElementById("info"); var data="class=cid"; xmlhttprequest.open("post",url,true); xmlhttprequest.setRequestHeader("content-type","application/x-www-form-urlencode"); xmlhttprequest.onreadystatechange=function(){ if(xmlhttprequest.readyState==4){ if(xmlhttprequest.status==200){ var info=xmlhttprequest.responseText; if(info==""){ show.value="你好，你查询的班级暂时还没有学生"; }else{ show.value=info; } } }else{ alert("error"); } } xmlhttprequest.send(data); } } </script> </head> <body> <div id="content"> <h1 id="请选择班级-select-nbsp-name-class-nbsp-id-class-nbsp-onchange-getMessage-option-nbsp-value-nbsp-selected-choose-nbsp-class-option-php-nbsp-foreach-info-nbsp-as-nbsp-v-option-nbsp-value-php-nbsp-echo-nbsp-v-cid-php-nbsp-echo-nbsp-v-cname-option-php-nbsp-select"> 请选择班级： <select name="class" id="class" onchange="getMessage()"> <option value="" selected>choose class</option> <?php foreach($info as $v){?> <option value="<?php echo $v['cid'];?>"><?php echo $v['cname'];?></option> <?php }?> </select> </h1> <div id="info"></div> </div> </body></html> 通过result.php查询数据库，并返回学生信息，代码如下：
<?php //header("content-type:text/html;charset=utf-8"); header("Content-type:text/xml;charset=utf-8"); header("Cache-Control:no-cache"); $class=$_POST['class']; mysql_connect("localhost","root","root") or die("数据库连接失败"); mysql_select_db("studentmanage"); mysql_query("set names utf-8"); $sql="select * from student where cid='$class'"; //$sql="select * from student where cid=3"; $rs=mysql_query($sql); $info=array(); while($row=mysql_fetch_assoc($rs)){ $info[]=$row; } $result="<table><tr><td>学号</td><td>姓名</td><td>住址</td><td>班级</td></tr>"; foreach($info as $v){ $result.="<tr><td>{$v['sid']}</td><td>{$v['name']}</td><td>{$v['address']}</td><td>{$v['cid']}</td></tr>"; } $result.="</table>"; echo $result;?> 我单独测试了下result.php，我确定没有问题，能从数据库中取出对应的信息，但是就是不能把取到的信息返回到index.php中
求各位帮帮忙，找找问题在哪
回复讨论(解决方案)
result.php
不能直接返回带输出内容，要用 json_encode 处理返回的数据
如：查表中的数据出来放到一数组中,然后：echo json_encode($arr);
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
result.php
不能直接返回带输出内容，要用 json_encode 处理返回的数据
如：查表中的数据出来放到一数组中,然后：echo json_encode($arr);
ajax不是有三种返回格式吗？分别为text, xml，json
我这里用的是text格式，我不知道这有什么错误
至少需要把 var show=document.getElementById("info"); 放在回调函数中
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
恩恩，你说的对(我Js基础不是很好==)，我按照您说的改了，然后按步骤调试了下：
<?php //从数据库把班级名称取出来放入下拉列表中 mysql_connect("localhost","root","root") or die("数据库连接失败"); mysql_select_db("studentmanage") or die("数据库不存在"); $sql="select * from class"; $rs=mysql_query($sql); $info=array(); while($row=mysql_fetch_assoc($rs)){ $info[]=$row; }?><!doctype html><html> <head> <title>查询学生信息</title> <meta http-equiv="content-type" content="text/html;charset=utf-8"/> <style type="text/css"> </style> <script type="text/javascript"> var xmlhttprequest;; function getMessage(){ if(window.ActiveXObject){ xmlhttprequest=new ActiveXObject("Microsoft.XMLHTTP"); }else{ xmlhttprequest=new XMLHttpRequest(); } if(xmlhttprequest){ var url="result.php"; var cid=document.getElementById("class").value; //window.alert(cid); var show=document.getElementById("info"); //window.alert(show); var data="class="+cid; //window.alert(data); xmlhttprequest.open("post",url,true); xmlhttprequest.setRequestHeader("content-type","application/x-www-form-urlencode"); xmlhttprequest.onreadystatechange=function(){ if(xmlhttprequest.readyState==4){ if(xmlhttprequest.status==200){ var info=xmlhttprequest.responseText; if(info==""){ show.innerHTML="你好，你查询的班级暂时还没有学生"; }else{ show.innerHTML=info; } } }else{ window.alert("error"); } } xmlhttprequest.send(data); } } </script> </head> <body> <div id="content"> <h1 id="请选择班级-select-nbsp-name-class-nbsp-id-class-nbsp-onchange-getMessage-option-nbsp-value-nbsp-selected-choose-nbsp-class-option-php-nbsp-foreach-info-nbsp-as-nbsp-v-option-nbsp-value-php-nbsp-echo-nbsp-v-cid-php-nbsp-echo-nbsp-v-cname-option-php-nbsp-select"> 请选择班级： <select name="class" id="class" onchange="getMessage()"> <option value="" selected>choose class</option> <?php foreach($info as $v){?> <option value="<?php echo $v['cid'];?>"><?php echo $v['cname'];?></option> <?php }?> </select> </h1> <div id="info">aaa</div> </div> </body></html> 下拉列表的值通过js取到了，我用window.alert(data)调试的时候没有问题
但是在result.php中通过$class=$_POST['class']取值的时候$class没有被赋值，因为我通过file_put_contents("D:/WWW/PHPDemo/log.log",$class."\r\n",FILE_APPEND);把$class的值保存在log.log文件中，但我运行后
打开时这个文件为空，且运行报错，说：Notice: Undefined index: class in D:\WWW\dbDesign\project\result.php on line 5
我想问下我的ajax通过POST传值的 方式有错误吗？？
<?php //header("content-type:text/html;charset=utf-8"); header("Content-type:text/xml;charset=utf-8"); header("Cache-Control:no-cache"); $class=$_POST['class']; file_put_contents("D:/WWW/PHPDemo/log.log",$class."\r\n",FILE_APPEND); mysql_connect("localhost","root","root") or die("数据库连接失败"); mysql_select_db("studentmanage"); mysql_query("set names utf-8"); $sql="select * from student where cid='$class'"; //$sql="select * from student where cid=3"; $rs=mysql_query($sql); $info=array(); while($row=mysql_fetch_assoc($rs)){ $info[]=$row; } $result="<table><tr><td>学号</td><td>姓名</td><td>住址</td><td>班级</td></tr>"; foreach($info as $v){ $result.="<tr><td>{$v['sid']}</td><td>{$v['name']}</td><td>{$v['address']}</td><td>{$v['cid']}</td></tr>"; } $result.="</table>"; echo $result;?> 问题就是没有把值传到result.php里面去，，，为什么呢，，一直没找到问题出在哪
xmlhttprequest.setRequestHeader("Content-Type","application/x-www-form-urlencode");
xmlhttprequest.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
最后少写了一个d
我能说我是一行行代码仔细看仔细看才找到错误的么=====
不知道有什么js调试工具能调试出这种错误，就不用这么费时间找BUG了。。。。
做程序员千万不能粗心，以此共勉
为什么不用jquery呢？
<script src="//code.jquery.com/jquery-1.11.0.min.js"></script><script type="text/javascript">var cid=document.getElementById("class").value;var show=document.getElementById("info");$.post("result.php", { class: cid }, function(info){ if(info==""){ show.innerHTML="你好，你查询的班级暂时还没有学生"; }else{ show.innerHTML=info; } },"text");</script> http://api.jquery.com/jQuery.post/
为什么不用jquery呢？
<script src="//code.jquery.com/jquery-1.11.0.min.js"></script><script type="text/javascript">var cid=document.getElementById("class").value;var show=document.getElementById("info");$.post("result.php", { class: cid }, function(info){ if(info==""){ show.innerHTML="你好，你查询的班级暂时还没有学生"; }else{ show.innerHTML=info; } },"text");</script> http://api.jquery.com/jQuery.post/
js才入门，我js基础很差的，我想把js基础打好点再用框架
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