2013-05-12 65 views
0

我試圖更新我的course_code到我的數據庫時複選框選中,但它失敗。我試圖從其他人那裏修改,但看起來像我犯了錯誤。誰能告訴我什麼是問題?更新使用複選框使用jquery

這裏是我的assigncourse.php

<?php require_once("../includes/session.php"); ?> 
<?php require_once("sessioncourse.php"); ?> 
<?php $course_codefac = $_SESSION['course_code'] ; ?> 
<?php confirm_logged_in(); ?> 
<?php require_once("../includes/connection.php") ?> 
<?php require_once("../includes/functions.php") ?> 


<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> 
<script> 
function Change(id) { 

$.ajax({ 
    type: "GET", 
    url: "updateassigncourse.php", 
    data: {"id": id}, 
    'success': function (response) { 
      console.log(response); 
      //TODO: use server response 
     } 
}); 
}; 
</script> 

<?php include("includes/header.php"); ?> 

    <div id="main"> 
    <div class="full_w"> 
    <?php 
      $querysel = "SELECT * FROM tblstudent ORDER BY student_id " ; 
     $resultsel = mysql_query($querysel, $connection); 

     echo "<h2><div class=\"h_title\">Please tick the student to join this 
".$course_codefac." course</div></h2>"; 
     echo "<table>"; 
     echo "<thead>"; 
     echo "<tr>"; 
     echo "<th scope=\"col\">Matric ID</th>"; 
     echo "<th scope=\"col\">Name</th>"; 
     echo "<th scope=\"col\">Assign</th>"; 
     echo "</tr>"; 
     echo "</thead>"; 


      while($rowsel = mysql_fetch_array($resultsel)){ 
     if($rowsel['course_code'] == NULL){ 
      $id = $rowsel['id']; 
        echo "<tr>"; 
        echo "<tr>"."<td class=\"align-center\">".$rowsel['student_id']." 
</td>"; 
        echo "<td class=\"align-center\">".$rowsel['name']."</td>"; 
     echo "<td class=\"align-center\">"; 
     echo "<input type=\"checkbox\" onchange=\"javascript: 
Change($id);\">"; 
echo "</td>"; 

     } 
     } 
     echo "</table>"; 




    ?> 
    </div> 
    </div> 
<?php include("includes/footer.php"); ?> 

那麼這裏就是我的updateassigncourse.php

<?php require_once("../includes/session.php"); ?> 
<?php require_once("sessioncourse.php"); ?> 
<?php $course_codeapp = $_SESSION['course_code'] ; ?> 
<?php confirm_logged_in(); ?> 
<?php require_once("../includes/connection.php") ?> 
<?php require_once("../includes/functions.php") ?> 

<?php 

$id = $_GET['id']; 
$course = $course_codeapp; 

$sql="UPDATE tblstudent set course_code = ". mysql_real_escape_string($course) 
." WHERE id = " .mysql_real_escape_string($id); 

$result = mysql_query($sql); 

?> 

回答

1

如果這是你確切的代碼,然後有一個缺少 ';'

$course = $course_codeapp 

它應該是

$course = $course_codeapp; 
+0

感謝通知,但在我的代碼我得到了放 「;」。也許剛剛被刪除。問題仍然沒有解決 – user2359110 2013-05-12 12:57:22

+0

'coursecode'的數據類型是什麼?如果它是一個varchar,你需要寫它'$ sql =「UPDATE tblstudent set course_code ='」。 mysql_real_escape_string($ course) 。「'WHERE id =」.mysql_real_escape_string($ id);' – draxxxeus 2013-05-12 13:13:36

+0

謝謝,問題解決= D – user2359110 2013-05-12 14:15:53