使用PHP使用下拉列表显示数据库中的数据

Question

我试图使用PHP基于http://www.w3schools.com/php/php_ajax_database.asp中的代码显示数据库中的数据。
而不是使用ID列，我使用我的演讲表中的TITLE列，它不起作用。没有显示任何内容（没有警告，没有任何内容）。
我该如何做这项工作？这里是我的代码

<script> 
    function showReport2(str) 
    { 
     if (str=="") 
     { 
      document.getElementById("txtHint").innerHTML=""; 
      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("txtHint").innerHTML=xmlhttp.responseText; 
      } 
     } 
     xmlhttp.open("GET","LectureReport.php?q="+str,true); 
     xmlhttp.send(); 
    } 
</script> 

我的HTML

<!-- lectures combo--> 
<label for = "LectTitle"> 
<span>Select Lecture:</span> 
&nbsp &nbsp &nbsp 
<select name="lectures" onchange="showReport2(this.value)"> 
    <option value="">----Select Lecture----</option> 
    <option value="1">Accounts</option> 
    <option value="Web Design">Web Design</option> 
    <option value="Database Administration">Database Administration</option> 
    <option value="Mathematics">Mathematics</option> 
    <option value="English">English</option> 
    <option value="French">French</option> 
    <option value="Mobile Device">Mobile Device</option> 
</select> 
</label> 

<br/> 
<hr> 

<div id="txtHint"> 
    <center> 
     Reports Will be displayed here 
    </center> 
</div> 

我的PHP脚本

<?php 
$q = intval($_GET['q']); 

$connect = mysqli_connect('localhost','root','','ewsd_coursework_db'); 
if (!$connect) 
{ 
    die('Could not connect: ' . mysqli_error($connect)); 
} 


$sql=" 
    SELECT 
     * 
    FROM 
     lecture 
    WHERE 
     LECTURE_NUMBER = '".$q."' 
    "; 

$result = mysqli_query($connect,$sql); 

echo " 
<table border='1'> 
<tr> 
    <th>Title</th> 
    <th>Overview</th> 
    <th>Room</th> 
    <th>Organiser</th> 
    <th>Presenter</th> 
    <th>Date</th> 
    <th>Time</th> 
</tr> 
"; 

while($row = mysqli_fetch_array($result)) 
{ 
    echo "<tr>"; 
    echo "<td>" . $row['TITLE'] . "</td>"; 
    echo "<td>" . $row['OVERVIEW'] . "</td>"; 
    echo "<td>" . $row['ROOM'] . "</td>"; 
    echo "<td>" . $row['ORGANISER'] . "</td>"; 
    echo "<td>" . $row['PRESENTER'] . "</td>"; 
    echo "<td>" . $row['DATE'] . "</td>"; 
    echo "<td>" . $row['TIME'] . "</td>"; 
    echo "</tr>"; 
} 
echo "</table>"; 

mysqli_close($connect); 
?> 

Answer 1

试试这个

while($row = $result->fetch_array()) 
{