0

我试图提出像谷歌这样的搜索建议......

我创建了一个表标签和一个单列标签，其中存储了一些标签，但问题是当我没有输入任何内容或数据库中的内容时，我得到未定义为返回

但是当我输入一些不在数据库中的东西时，我很抱歉[这是正确的]

我正在使用 Ajax

我的PHP代码

<?php

header('Content-Type: text/xml');

echo '<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>';

echo '<response>';

        $searchvalue=$_GET['searchvalue'];

        $con=mysqli_connect("localhost","root","","myweb") or die("error connecting db");
        $sqlresult = mysqli_query($con,"SELECT tag FROM tags where tag like '".$searchvalue."%'");
        $suggests="";
        while($row = mysqli_fetch_array($sqlresult))
        {
            $suggests=$suggests.$row.',';
        }

        //$suggarray=explode(",",$suggests);

        if(strlen($suggests)>0)
        {
            echo "found";
        }
        else
        {
            echo "sorry! for ".$searchvalue;
        }

        mysqli_close($con);
    echo '</response>';
?>

用于 ajax 的 JavaScript

// JavaScript for search ajax

var xmlHttp= createXmlHttpRequestObject();

function createXmlHttpRequestObject(){
    //alert("create obj");
   var xmlHttp;

   if(window.ActiveXObject){
      try{
         xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
         }catch(e){
            xmlHttp =false;
            }
      }else{
         try{
            xmlHttp= new XMLHttpRequest();
            }catch(e){
               xmlHttp =false;
               }
         }
      if(!xmlHttp)
            alert("cant create that object hoss!");
      else
            return xmlHttp;
   }

function searchprocess(){ 
    //alert("start");
    if(xmlHttp.readyState==0 || xmlHttp.readyState==4){
      searchtext=encodeURIComponent(document.getElementById("searchvalue").value );
      xmlHttp.open("GET", "pages/search_suggestions.php?searchvalue="+searchtext, true);
      xmlHttp.onreadystatechange = handleServerResponse;
      xmlHttp.send(null);
      }else{
         setTimeout('searchprocess()', 1000);
         }
   }


function handleServerResponse(){
    //alert("handle");
   if(xmlHttp.readyState==4){
            //alert("handle ready");
            if(xmlHttp.status==200 || xmlHttp.status==304){
            //alert("inside");
               xmlResponse=xmlHttp.responseXML;
               xmlDocumentElement=xmlResponse.documentElement;
              message=xmlDocumentElement.firstChild.data;
               document.getElementById("tempo").innerHTML=message;
               setTimeout('searchprocess()', 1000);
         }else{
            alert("Something went wrong!");
            }
      }
   }


// end of JavaScript for search ajax
4

2 回答 2

1

由于您的 php 脚本由 AJAX 调用，而不是单独的回显，将它们连接成一个字符串并在最后回显一次：

header('Content-Type: text/xml');
$resp = '<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>';
$resp .= '<response>';

..and so on... until

$resp .= '</response>';

然后echo $response;

于 2013-04-26T03:21:12.627 回答
0 
   $sqlresult = mysqli_query($con,"SELECT tag FROM tags where tag like '".$searchvalue."%'");
    $suggests="";
    while($row = mysqli_fetch_array($sqlresult))
    {
$comma=$suggests?',':'';
        $suggests.=$comma.$row['tag']; 
    }
于 2013-04-26T03:26:56.260 回答