PHP and AJAX Suggest

PHP and AJAX Suggest

這個suggest是仿google的效果了,google總是走在世界的技術的前面啊,下面我們也來看看實例吧.

先看看html文件.

<form> 
First Name:
<input type="text" id="txt1"
onkeyup="showHint(this.value)">
</form>

<p>Suggestions: <span id="txtHint"></span></p>

下面用js文件的ajax文件來處理數據.

var xmlHttp;

function showHint(str)
{
if (str.length==0)
  { 
  document.getElementById("txtHint").innerHTML="";
  return;
  }
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
  {
  alert ("Browser does not support HTTP Request");
  return;
  } 
var url="gethint.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
} 

function stateChanged() 
{ 
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
 { 
 document.getElementById("txtHint").innerHTML=xmlHttp.responseText;
 } 
}
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;
}
 
這個文件只是返回php處理的數據,而我們的php文件如下.
<?php
// 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[]="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; $i<count($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;
?>
好了我們的確suggest效果就完成了.