通过ajax POST传值，但返回为空-PHP Tutorial-php.cn

Home

Backend Development

PHP Tutorial

通过ajax POST传值，但返回为空

WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWB

Jun 23, 2016 pm 01:58 PM

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

Statement

The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn

Working with Flash Session Data in LaravelMar 12, 2025 pm 05:08 PM

Laravel simplifies handling temporary session data using its intuitive flash methods. This is perfect for displaying brief messages, alerts, or notifications within your application. Data persists only for the subsequent request by default: $request-

cURL in PHP: How to Use the PHP cURL Extension in REST APIsMar 14, 2025 am 11:42 AM

The PHP Client URL (cURL) extension is a powerful tool for developers, enabling seamless interaction with remote servers and REST APIs. By leveraging libcurl, a well-respected multi-protocol file transfer library, PHP cURL facilitates efficient execution of various network protocols, including HTTP, HTTPS, and FTP. This extension offers granular control over HTTP requests, supports multiple concurrent operations, and provides built-in security features.

Simplified HTTP Response Mocking in Laravel TestsMar 12, 2025 pm 05:09 PM

Laravel provides concise HTTP response simulation syntax, simplifying HTTP interaction testing. This approach significantly reduces code redundancy while making your test simulation more intuitive. The basic implementation provides a variety of response type shortcuts: use Illuminate\Support\Facades\Http; Http::fake([ 'google.com' => 'Hello World', 'github.com' => ['foo' => 'bar'], 'forge.laravel.com' =>

12 Best PHP Chat Scripts on CodeCanyonMar 13, 2025 pm 12:08 PM

Do you want to provide real-time, instant solutions to your customers' most pressing problems? Live chat lets you have real-time conversations with customers and resolve their problems instantly. It allows you to provide faster service to your custom

Explain the concept of late static binding in PHP.Mar 21, 2025 pm 01:33 PM

Article discusses late static binding (LSB) in PHP, introduced in PHP 5.3, allowing runtime resolution of static method calls for more flexible inheritance.Main issue: LSB vs. traditional polymorphism; LSB's practical applications and potential perfo

PHP Logging: Best Practices for PHP Log AnalysisMar 10, 2025 pm 02:32 PM

PHP logging is essential for monitoring and debugging web applications, as well as capturing critical events, errors, and runtime behavior. It provides valuable insights into system performance, helps identify issues, and supports faster troubleshoot

Discover File Downloads in Laravel with Storage::downloadMar 06, 2025 am 02:22 AM

The Storage::download method of the Laravel framework provides a concise API for safely handling file downloads while managing abstractions of file storage. Here is an example of using Storage::download() in the example controller:

HTTP Method Verification in LaravelMar 05, 2025 pm 04:14 PM

Laravel simplifies HTTP verb handling in incoming requests, streamlining diverse operation management within your applications. The method() and isMethod() methods efficiently identify and validate request types. This feature is crucial for building

See all articles