我正在創建一個社交網站。我在search.php
的文本框中搜索存儲在數據庫中的場所。當用戶開始鍵入場地名稱時,數據庫中存儲的場地名稱列表應該是加載到div venuesearch
。這工作正常。我的問題是,我需要在div venuesearch
可點擊內的行,以便當用戶單擊一個特定的行時,該值應該來到文本框。連續點擊
的search.php
<script language="javascript">
function showData(str)
{
if (str.length==0)
{
document.getElementById("venuesearch").innerHTML="";
document.getElementById("venuesearch").style.border="0px";
return;
}
var venue_data = document.getElementById("venue").value;
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("venuesearch").innerHTML=xmlhttp.responseText;
document.getElementById("venuesearch").style.border="1px solid #A5ACB2";
document.getElementById("venuesearch").style.zIndex = "100";
}
}
xmlhttp.open("GET","aj_search.php?venue="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<input id="venue" name="venue" type="text" onkeyup="showData(this.value)" value="Tag venue name"/>
<div id="venuesearch">
</div>
</body>
aj_search.php
<?php
$dbhandle=mysql_connect("localhost","root","")or die("Unable to connect");
$select=mysql_select_db("scenekey",$dbhandle) or die("Unable to connect");
if(isset($_GET['venue']))
{
$venue_name = $_GET['venue'];
}
$hint='';
$query_get_venue = "SELECT * from sk_accounts WHERE acnt_member_class='venue' and acnt_fname LIKE '".$venue_name."%'";
$result_query_get_venue = mysql_query($query_get_venue);
$row_count = mysql_num_rows($result_query_get_venue);
if($row_count > 0)
{
$hint = "<table>";
while($row = mysql_fetch_array($result_query_get_venue))
{
$act_fname = $row['acnt_fname'];
$act_lname = $row['acnt_lname'];
$act_name = $act_fname . ' ' . $act_lname;
$hint.= "<tr><td>".$act_name."</td></tr>";
}
$hint .= "</table>";
}
if ($hint == "")
{
$response="no suggestion";
}
else
{
$response=$hint;
}
//output the response
echo $response;
?>
你解決你的問題? – Vimalnath