Question

我试图从表单中获取输入并删除表格数据，然后显示更新的表格，但我得到一个空白页面，我不知道这可能是什么问题，任何帮助都会在这里得到关注＃39;我的代码：

  <html>
        <body>
<?php
 $mysqli = new mysqli("xxxxx", "xxxxxx", "xxxxx", "xxxxxx");

 /* check connection */ 
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
 //----------------------------------------------------------------------------------//
$name = $_POST['Car_ID'];

if ($stmt = $mysqli->prepare("delete from CARS where name=?")) {

    // Bind the variable to the parameter as a string. 
    $stmt->bind_param("s", $name);

    // Execute the statement.
    $stmt->execute();

 echo "Deleted data successfully\n";

    // Close the prepared statement.


  $mysqli->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $result = $db->prepare("SELECT id, Doors, TRANSMISSION, Fuel_type, Engine_Size, Total FROM CARS");
        $result->execute();
        while ($row = $result->fetch(PDO::FETCH_ASSOC)){
            $doors=$row["Doors"];
            $engine=$row["Engine_Size"];
            $total=$row["Total"];
            $trans=$row["Transmission"];
            }
         ?>

        <table>
        <tr>
        <td><?php echo $doors; ?></td>
        <td><?php echo $engine; ?></td>
        <td><?php echo $total; ?></td>
        <td><?php echo $trans; ?></td>

        </tr>
  <?php } ?>
        </table>
        </body>
        </html>

Answer 1

您正在将mysqli与PDO混合

$mysqli->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

并且您正在通过$db（{em>理论上）应该$mysqli

$result = $db->prepare("SELECT id, Doors, TRANSMISSION, Fuel_type, Engine_Size, Total, DATE_INITIATED, AGE, PARTNO, QTY, DESCRIPTION, LOC  FROM CARS");

理论上，它应该是

$result = $mysqli->prepare("SELECT id, Doors, TRANSMISSION, Fuel_type, Engine_Size, Total, DATE_INITIATED, AGE, PARTNO, QTY, DESCRIPTION, LOC  FROM CARS");

因为您的数据库连接是：

$mysqli = new mysqli("xxxxx", "xxxxxx", "xxxxx", "xxxxxx");

但是，您的数据库连接应该类似于：

$mysql_hostname = 'xxx';
$mysql_username = 'xxx';
$mysql_password = 'xxx';
$mysql_dbname = 'xxx';

$db = new PDO("mysql:host=$mysql_hostname;dbname=$mysql_dbname", $mysql_username, $mysql_password);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

你不能这样做：

$stmt->bind_param("s", $name);

您是否记得使用PDO？（或者是你？）接受你的选择，是mysqli还是PDO？

你最有可能想做的事：

// $stmt = $db->prepare("delete from CARS where name=:value");

if ($stmt = $db->prepare("delete from CARS where name=:value")) {    
// Bind variables to your statement
$stmt->bindParam(':value', $name);

... }

Answer 2

你启用了error_reporting吗？

error_reporting(E_ALL);
ini_set("display_errors","on");

将其添加到PHP文件的头部。

Answer 3

很棒points made by @fred，当您最终确定要使用哪些数据库扩展时，你仍然需要修改输出表数据的方式，你应该在循环外面有表标签，和你应该在循环内打印表行，注意额外的括号。

这样的事情会起作用：

<html>
<body>
    <table>
<?php

   //Fetch your data
   //.....
   //....
    while ($row = $result->fetch(PDO::FETCH_ASSOC)){
    echo"<tr>
            <td>$row['Doors']</td>
            <td>$row['Engine_Size']</td>
            <td>$row['Total']</td>
            <td>$row['Transmission']</td>
        </tr>";
    }


?>
    </table>
</body>
</html>

带有PHP脚本的空白页面

3 个答案: