这个PHP Ajax实时搜索的XML结构 - XML structure for this PHP Ajax Live Search

XML structure for this PHP Ajax Live Search

嗨，伙计们，我看到这个PHP Ajax Live搜索示例在这里。我试着解决这个问题。但是xml结构失败了。你能帮我建立最好的xml结构为这段代码_?

HTML:

<html>
<head>
<script>
function showResult(str) {
  if (str.length==0) { 
    document.getElementById("livesearch").innerHTML="";
    document.getElementById("livesearch").style.border="0px";
    return;
  }
  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp=new XMLHttpRequest();
  } else {  // code for IE6, IE5
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {
      document.getElementById("livesearch").innerHTML=xmlhttp.responseText;
      document.getElementById("livesearch").style.border="1px solid #A5ACB2";
    }
  }
  xmlhttp.open("GET","livesearch.php?q="+str,true);
  xmlhttp.send();
}
</script>
</head>
<body>
<form>
<input type="text" size="30" onkeyup="showResult(this.value)">
<div id="livesearch"></div>
</form>
</body>
</html>

PHP:

<?php
$xmlDoc=new DOMDocument();
$xmlDoc->load("links.xml");
$x=$xmlDoc->getElementsByTagName('link');
//get the q parameter from URL
$q=$_GET["q"];
//lookup all links from the xml file if length of q>0
if (strlen($q)>0) {
  $hint="";
  for($i=0; $i<($x->length); $i++) {
    $y=$x->item($i)->getElementsByTagName('title');
    $z=$x->item($i)->getElementsByTagName('url');
    if ($y->item(0)->nodeType==1) {
      //find a link matching the search text
      if (stristr($y->item(0)->childNodes->item(0)->nodeValue,$q)) {
        if ($hint=="") {
          $hint="<a href='" . 
          $z->item(0)->childNodes->item(0)->nodeValue . 
          "' target='_blank'>" . 
          $y->item(0)->childNodes->item(0)->nodeValue . "</a>";
        } else {
          $hint=$hint . "<br /><a href='" . 
          $z->item(0)->childNodes->item(0)->nodeValue . 
          "' target='_blank'>" . 
          $y->item(0)->childNodes->item(0)->nodeValue . "</a>";
        }
      }
    }
  }
}
// Set output to "no suggestion" if no hint was found
// or to the correct values
if ($hint=="") {
  $response="no suggestion";
} else {
  $response=$hint;
}
//output the response
echo $response;
?>

我的问题是，什么是最好的XML结构，这工作正确?

我试过这样做。但是它只显示一个结果。

<link>
<title>Spaksu</title><url>spaksu.com</url>
<title>Blogaps</title><url>blogaps.com</url>
</link>

? ?

谢谢. .

Try 
<link>
<title>Spaksu</title>
<url>spaksu.com</url>
</link>
<link>
<title>Blogaps</title>
<url>blogaps.com</url>
</link>