所以我复制/粘贴了一些用于实时搜索的代码(来自w3schools的代码),当用户开始输入时,会出现一个带有一些超链接字的下拉菜单。当用户单击这些超链接时,它将打开一个新页面,其中包含他们单击的特定项目。目前,这些超链接都是.html,我假设以下行默认在.html页面中打开这些页面:
$z=$x->item($i)->getElementsByTagName('url');
我的问题是:如何将其链接到.php页面呢?
以下是显示结果的JS代码:
<script type="text/javascript">
function showResult(str)
{
if (str.length==0)
{
document.getElementById("search").innerHTML="";
document.getElementById("search").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("search").innerHTML=xmlhttp.responseText;
document.getElementById("search").style.border="1px solid #A5ACB2";
}
}
xmlhttp.open("GET","search2.php?q="+str,true);
xmlhttp.send();
}
</script>
这是用于搜索的PHP:
<?php
$xmlDoc=new DOMDocument();
$xmlDoc->load("catalogue.xml");
$x=$xmlDoc->getElementsByTagName('flower');
//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('commonname');
$z=$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'>" .
$y->item(0)->childNodes->item(0)->nodeValue . "</a>";
//$hint= $y->item(0)->childNodes->item(0)->nodeValue. "<br />";
}
else
{
$hint=$hint . "<br /><a href='" .
$z->item(0)->childNodes->item(0)->nodeValue .
"' target='_blank'>" .
$y->item(0)->childNodes->item(0)->nodeValue . "</a>";
//$hint= $hint.$y->item(0)->childNodes->item(0)->nodeValue. "<br />";
}
}
}
}
}
// 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;
?>