php+ajax 实现输入读取数据库显示匹配信息

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废话不多说了，直接跟大家贴代码了

dropbox_index.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>主搜索引擎下拉框自动显示数据</title>
</head>
<script src="/UploadFiles/2021-04-02/dropbox.js">

dropbox.js    ajax     


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+""+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;
    }


dropbox.css
.txt{ height:22px; border:1px solid #cdcdcd; width:220px;border-right:none;} 

.btn{ background-color:#FFF; border:1px solid #CDCDCD;height:26px; width:70px;}
responsepage.php


<"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'] . "<br />";
         //echo "lastname:" . $row['LastName'] . "<br />";
         //echo "age:" . $row['Age'] . "<br />";
       }
       //echo $row['FirstName'] . "<br />";
     }
    mysql_close($con);
   }
"text-align: center">


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