Question

我尝试根据用户点击的当前主题（主题称为会话）从测验列表中显示测验。会话有练习，每次练习我都希望选择参加特定的课程测验（一个课程有5个练习和1个测验）。点击quiz按钮后，我尝试使用当前$session_id显示测验标题和作者，并将其与session_id表格中的quiz_list相关联在数据库中。

没有显示任何内容，我收到的错误如下：

注意：未定义的索引：第8行的C：\ xampp \ htdocs \ project \ quiz.php中的session_id

警告：mysqli_fetch_assoc（）要求参数1为mysqli_result，第13行的C：\ xampp \ htdocs \ Project \ quiz.php中给出布尔值

exercise.php：

<?php 
    //build up array as iterate through while loop
   $sessions_quiz_array = array();
   $get_quiz = mysqli_query($con, "SELECT * FROM `quiz_list` WHERE `session_id` = $session_id");

   //fetches result row as an associative array
   while ($row_quizes = mysqli_fetch_assoc($get_quiz)) {
   $sessions_quiz_array[] = $row_quizes;

   }

   //sessions_quiz_array is the associative array being looped 
   foreach ($sessions_quiz_array as $key => $value) {
      $quiz_id = $value['id'];
      $quiz_title = $value['quiz_title'];
      $quiz_author = $value['quiz_author'];
      ?>

    <a id='button' href="quiz.php?id=<?php echo $quiz_id; ?>"> Quiz</a>


            <?php  } ?>

quiz.php

   <?php 
   session_start(); 
   include("includes/database.php");
   ?>
    <?php 
        //gets current
        $session_id = $_GET['session_id'];

    $get_quiz = "SELECT * FROM `quiz_list` WHERE `session_id` = $session_id";

        $run_quiz = mysqli_query($con, $get_quiz);
        $quiz_info = mysqli_fetch_assoc($run_quiz);

     ?>
      <hr>
      <div>

        <h1><?php echo $quiz_info["id"]; ?> </h1>
        <br>
        <p><?php echo $quiz_info["quiz_title"]; ?></p>
        <br>
        <p><strong><?php echo $quiz_info["quiz_author"]; ?></strong></p>


      </div>
      <hr>

      <a href="interactive_training.php" class="button previous">&laquo; Back to Training</a>

非常感谢任何帮助。

Answer 1

在您的文件exercise.php中，您的GET请求为ID而session_id附近quiz.php?id=

您需要在quiz.php文件中进行以下更改。

从

改变

$session_id = $_GET['session_id'];

要

$session_id = $_GET['id'];

Addiotionally你可以在执行整个页面之前添加检查，如

if(!isset($session_id))
{
    exit('Session ID is missing');
}

Answer 2

您的quiz.php变为： -

替换此

 $session_id = $_GET['session_id'];

与

 $session_id = $_GET['id'];

通过php和html显示测验

2 个答案: