Question

我正在尝试删除和输入而不刷新，这是我到目前为止但由于某种原因无效。

信息显示在表格中，工作正常，问题是当我尝试删除没有发生的事情时。

这是我的courseData.php

<script type="text/javascript">
function deleteC(){
    var url='deleteCourse.php';
    var id = $F($( <?php ' . $row['courseid'] . ' ?>));
    var myAjax = new Ajax.Updater(url,{method: 'get', paramaters: 'id='+id'});
<?php


echo "<table width='100%'>";
echo "<tr> 
      <th>Course name</th> 
      <th>Delete</th>
      <th>Edit</th>
      </tr>";
?>
    <?php foreach($rows as $row):  
    echo "<tr>"; 
    echo '<td><a href="#">' . htmlentities($row['coursename'], ENT_QUOTES, 'UTF-8') . '</a></td>';
    echo '<td><a href="finalphp/deleteCourse.php?id=' . $row['courseid'] . '" onclick="deleteC()");"><font color="#e70404"> Delete </font> </a></td>';
    echo '<td><a class="delete" id="'.$row["courseid"].'">Delette</a></td>';
    echo "</tr> ";
    endforeach; 
echo "</table>";

这是我的deleteCourse.php

<?php 
require("connect.php");
if (isset($_GET['id']) && is_numeric($_GET['id']))
{ 
    $id = $_GET['id'];

    echo"$id"; 
    $con=mysqli_connect("localhost","root","","independentstudyclass");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    mysqli_query($con,"DELETE FROM courses WHERE courseid=$id");

    mysqli_close($con);

    }

?>

Answer 1

请注意：

var myAjax = new Ajax.Updater(url,{method: 'get', paramaters: 'id='+id'});

我认为，id之后的最后一句话不应该存在。

删除而不刷新

1 个答案: