php+ajax 实现输入读取数据库显示匹配信息，phpajax_PHP教程

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="responsepage.php" url=url+"?q="+str url=url+"&sid="+Math.random() xmlHttp.onreadystatechange=stateChanged xmlHttp.open("GET",url,true) xmlHttp.send(null) } //设置回调函数 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; }

<?php //get the q parameter from URL $q=$_GET["q"]; //全部小写化 $q=strtolower($q); //非空验证 if(isset($q) && $q != '') { $con = mysql_connect("localhost","root","lifu"); if(!$con) { die('Could not connect: ' .mysql_error()); } mysql_select_db("my_db",$con); $sql = "select * from persons"; //$sql="SELECT FirstName FROM Persons where Firstname like '%$q%'"; $result =mysql_query($sql,$con); while($row = mysql_fetch_array($result)) { //匹配判断 if(stristr(strtolower($row['FirstName']),$q)) { //echo "-----------------Persons-----------------"; echo /* "firstname:" .*/ $row['FirstName'] . "
"; //echo "lastname:" . $row['LastName'] . "
"; //echo "age:" . $row['Age'] . "
"; } //echo $row['FirstName'] . "
"; } mysql_close($con); } ?>