使用PHP MySQLi的HTML表格

Question

我试图用PHP MySQLi向表格（HTML表格）显示一些信息。我目前在我创建的网页上发生错误。该图像显示了我的页面图像和错误。

<?PHP 
    $servername = "localhost"; 
    $username = "root"; 
    $password = ""; 
    $dbname = "ip_site_database_project"; 

    $conn = new mysqli($servername, $username, $password, $dbname); 

    if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
    } 
    $sql = "SELECT * FROM site"; 
    $result = mysqli_query($conn, $sql); 

    echo "<table border='1' align ='center'>"; 
    echo "<tr>"; 
     echo "<td>Site ID</td>"; 
     echo "<td>Site Name</td>"; 
     echo "<td>Site Type</td>"; 
     echo "<td>Description</td>"; 
     echo "<td>Site Dial Code</td>"; 
     echo "<td>Site Number Range (Start)</td>"; 
     echo "<td>Site Number Range (End)</td>"; 
     echo "<td>Site Active Since</td>"; 
     echo "<td>Site Ceased Date</td>"; 
     echo "</tr>"; 
    while($rowitem = mysqli_fetch_array($results)) 
    { 
     echo"<tr>"; 
     echo "<td>" . $rowitem[Site_ID] . "</td>"; 
     echo "<td>" . $rowitem[Site_Name] . "</td>"; 
     echo "<td>" . $rowitem[Site_Type] . "/td>"; 
     echo "<td>" . $rowitem[Description] . "</td>"; 
     echo "<td>" . $rowitem[Site_Dial_Code] . "/td>"; 
     echo "<td>" . $rowitem[Site_Number_Range_Start] . "</td>"; 
     echo "<td>" . $rowitem[Site_Number_Range_End] . "/td>"; 
     echo "<td>" . $rowitem[Site_Active_Since] . "</td>"; 
     echo "<td>" . $rowitem[Site_Ceased_Date] . "/td>"; 
     echo "</tr>"; 
    } 
    echo"</table>"; 

    $conn->close(); 

    ?> 

Answer 1

更改此：while($rowitem = mysqli_fetch_array($results))这样：while($rowitem = mysqli_fetch_array($result))