2014-03-30 80 views
0

我想刪除和輸入沒有刷新,這是我迄今爲止,但由於某種原因是行不通的。刪除沒有刷新

信息顯示在表格中,工作正常,問題是當我嘗試刪除沒有發生的事情時。

這是我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>"; 

,這是我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); 

    } 

?> 
+0

第一步,是PHP甚至叫? –

+0

我使用'url'將php傳遞到myAjax var中 – user3424633

+0

您需要關閉腳本標記。另外,由於'var id'部分不在for循環中,所以它不會有任何值。 –

回答

0

注意這一點:

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

最後報價剛過ID不應該在那裏,我想。