我想写一个ajax脚本，但我想不通。
输入姓名后，必须运行一个查询来搜索此人的id。
jquery/html

<script type="text/javascript">
 $(document).ready(function(){
 	$("searchName").change(function(){
 		var name = $("searchName").val();
 		$.post("searchid.php", {
 			naam: name
 		}, function(data, status) {
 			$("#test").html(data);
 			});
 		});
 	});
</script>

<table>
<div class="auto-widget">
	<tr>
    	<td><lable>Naam</lable></td>
    	<td><input name="serarchName" class="tagert" type="text" id="searchName"/></td>
    <tr>
    <tr>
    	<td><lable>id</lable></td>
    	<td><input type="text" id="tmnlId"/></td>
    <tr>	
</div>
</table>

<p id="test"></p>

PHP

//set $server, $user, $password and $database_name, then establish the connection:
    $conn = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
    if ($conn->connect_error) {
        exit("FALSE");
    }

    $naam = $_POST["naam"];
    $query=mysqli_query($conn,"SELECT * FROM Medewerkers WHERE naam ='".$naam."'");
    $row=mysqli_fetch_array($query);
    echo $row['id'];

    ?>
    enter code here