这是我编写的AJAX代码:
这是第27页代码:27.PHP
<script language=javascript>
function InitAjax()
{
var ajax=false;
try
{
ajax = new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
ajax = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (E)
{
ajax = false;
}
}
if (!ajax && typeof XMLHttpRequest!='undefined')
{
ajax = new XMLHttpRequest();
}
return ajax;
}
var ajax = InitAjax();
ajax.open("POST", "28.php?id=3", true);
ajax.send(null);
</script>
这是第28页代码:28.php
<?php
$conn = @ mysql_connect("localhost", "root", "") or die("数据库链接错误");
mysql_select_db("test", $conn);
$id = $_POST['id'];
$sql="INSERT INTO te (id) VALUES ($id)"; //te 这里是 test下一个表只有一个字段id
mysql_query($sql);
?>
我只运行27.php时28.php好像根本没作用,数据库中查不到添加的信息。这代码可能不规范,如果不好改,请求有没有类似可用的实例,给小弟研读一下,呵呵!
这是第27页代码:27.PHP
<script language=javascript>
function InitAjax()
{
var ajax=false;
try
{
ajax = new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
ajax = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (E)
{
ajax = false;
}
}
if (!ajax && typeof XMLHttpRequest!='undefined')
{
ajax = new XMLHttpRequest();
}
return ajax;
}
var ajax = InitAjax();
ajax.open("POST", "28.php?id=3", true);
ajax.send(null);
</script>
这是第28页代码:28.php
<?php
$conn = @ mysql_connect("localhost", "root", "") or die("数据库链接错误");
mysql_select_db("test", $conn);
$id = $_POST['id'];
$sql="INSERT INTO te (id) VALUES ($id)"; //te 这里是 test下一个表只有一个字段id
mysql_query($sql);
?>
我只运行27.php时28.php好像根本没作用,数据库中查不到添加的信息。这代码可能不规范,如果不好改,请求有没有类似可用的实例,给小弟研读一下,呵呵!
ajax.open("POST", "28.php", true);//这句开始
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");//POST方式,需要
ajax.send("id=3");//传参放这里
$id = $_POST['id'];
改为$id = $_GET['id'];
然后你先不要post或get id。自己设一个看看有没有返回值。
然后再从27.php传值过去看看
这是html:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>在此处插入标题</title>
<script type="text/javascript">
var xmlHttp=null;
function showhit(str){
if(str.length==0){
document.getElementById("special").innerHTML="";
return;
}
xmlHttp=GetXMLHttpObject();
if(xmlHttp==null){
alert("Your broswer does not support HTTP Request!");
return;
}
var url="special.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=statuChange;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
function statuChange(){
if(xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
document.getElementById("special").innerHTML=xmlHttp.responseText;
}
} function GetXMLHttpObject(){
try{
xmlHttp=new XMLHttpRequest();
}catch(e){
try{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}catch (e){
xmlHttp=new ActiveXObject("Microft.XMLHTTP");
}
}
return xmlHttp;
}
</script>
</head>
<body>
<form>
<label style="font-size: 20px;">username:</label>
<input type="text" id="nameTabel" onkeyup="showhit(this.value)"/><br>
<label>suggestion:</label><span id="special" style="color:red;font-size:20px;"></span>
</form>
</body>
</html>下面是处理php:<?php$a[]="yulongxinag";
$a[]="sunxiaomeng";
$a[]="yangcheng";
$a[]="lanzhiyu";
$a[]="zhengwenbo";
$a[]="luoshicong";
$a[]="liuxingping";
$a[]="zhangjinxiong";
$a[]="wangmaolin";
$a[]="lianfeng";
$q=$_GET["q"];//lookup all hints from array if length of q>0
if (strlen($q) > 0)
{
$hint="";
for($i=0; $i<count($a); $i++)
{
if (strtolower($q)==strtolower(substr($a[$i],0,strlen($q))))
{
if ($hint=="")
{
$hint=$a[$i];
}
else
{
$hint=$hint." , ".$a[$i];
}
}
}
}//Set output to "no suggestion" if no hint were found
//or to the correct values
if ($hint == "")
{
$response="no suggestion";
}
else
{
$response=$hint;
}//output the response
echo $response;
?>不知道对你有没有帮助
ajax.send(null);
这里对吗?post应该在send里传值吧。
ajax.open("POST", "28.php", true);
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
ajax.send("id=3");