php的ajax简单实例
当输入j后,会触发ajax效果,从后台获取相应的名字中带有j的数据,并展示在suggestions中。
代码实现如下:
实现ajax需要三个文件,一个是html的表单文件,一个是js的核心文件,一个是php的后台文件。
下面的是html文件,当键盘按下时触发showHint方法,在showHint方法中会有ajax的核心内容,实例化,获取地址,获取数据并展示等等。
复制代码 代码如下:
Suggestions:
下面是js的内容clienthint.js。
复制代码 代码如下:
var xmlHttp
function showHint(str)
{
if (str.length==0)
{
document.getElementById("txtHint").innerHTML=""
return
}
//获取xmlHttpObject对象,如果为空,提示浏览器不支持ajax
xmlHttp=GetXmlHttpObject()
if (xmlHttp==null)
{
alert ("Browser does not support HTTP Request")
return
}
//获取url
var url="gethint.php"
url=url+"?q="+str
url=url+"&sid="+Math.random()
//回调函数,执行动作
xmlHttp.onreadystatechange=stateChanged
//open
xmlHttp.open("GET",url,true)
xmlHttp.send(null)
}
function stateChanged()
{
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
//将获取的信息插入到txtHint中
document.getElementById("txtHint").innerHTML=xmlHttp.responseText
}
}
//获取xml对象
function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
下面是php的内容。根据ajax对象传入的参数,获取相应的数据。
复制代码 代码如下:
// Fill up array with names
$a[]="Anna";
$a[]="Brittany";
$a[]="Cinderella";
$a[]="Diana";
$a[]="Eva";
$a[]="Fiona";
$a[]="Gunda";
$a[]="Hege";
$a[]="Inga";
$a[]="Johanna";
$a[]="Jiqing";
$a[]="Kitty";
$a[]="Linda";
$a[]="Nina";
$a[]="Ophelia";
$a[]="Petunia";
$a[]="Amanda";
$a[]="Raquel";
$a[]="Cindy";
$a[]="Doris";
$a[]="Eve";
$a[]="Evita";
$a[]="Sunniva";
$a[]="Tove";
$a[]="Unni";
$a[]="Violet";
$a[]="Liza";
$a[]="Elizabeth";
$a[]="Ellen";
$a[]="Wenche";
$a[]="Vicky";
//get the q parameter from URL
$q=$_GET["q"];
//lookup all hints from array if length of q>0
if (strlen($q) > 0)
{
$hint="";
for($i=0; $i
if (strtolower($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;
?>

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