它不会从下拉菜单中显示正确的选项详细信息

Question

我正在尝试检索从课程下拉模块中选择的课程的详细信息，然后显示一个模块下拉菜单，其中列出了属于该课程的所有模块。

我遇到的问题是，可以说我的课程下拉菜单中有这两个选项：

INFO101 - Business
INFO102 - ICT

出于某种奇怪的原因，每次我从下拉菜单中选择顶部选项（INFO101）并单击提交按钮时，它总是显示其他课程详细信息（INFO102），从而显示属于该课程的模块和不是其他课程。

我的问题是，当我从下拉菜单中提交（INFO101）选项时，为什么会显示其他课程的信息？

下面是mysqli代码

     $sql = "SELECT CourseId, CourseName FROM Course"; 

$sqlstmt=$mysqli->prepare($sql);

$sqlstmt->execute(); 

$sqlstmt->bind_result($dbCourseId, $dbCourseName);

$courses = array(); // easier if you don't use generic names for data 

$courseHTML = "";  
$courseHTML .= '<select name="courses" id="coursesDrop">'.PHP_EOL; 
$courseHTML .= '<option value="">Please Select</option>'.PHP_EOL;  

while($sqlstmt->fetch()) 
{ 
$course = $dbCourseId;
$coursename = $dbCourseName; 
$courseHTML .= '<option value="'.$course.'">' . $course . ' - ' . $coursename . '</option>'.PHP_EOL;  
} 

$courseHTML .= '</select>'; 
$courseHTML .= '</form>'; 

?>

<?php
include('noscript.php');
?>

<h1>CREATING A NEW ASSESSMENT</h1>

<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post">
<table>
<tr>
<th>Course: <?php echo $courseHTML; ?><input id="courseSubmit" type="submit" value="Submit" name="submit" /></th>
</tr>
</table>
</form>

<?php
if (isset($_POST['submit'])) {

$submittedCourseId = (isset($_POST['courses']));

$query = "
SELECT cm.CourseId, cm.ModuleId, 
c.CourseName,
m.ModuleName
FROM Course c
INNER JOIN Course_Module cm ON c.CourseId = cm.CourseId
JOIN Module m ON cm.ModuleId = m.ModuleId
WHERE
(c.CourseId = ?)
ORDER BY c.CourseName, m.ModuleId
";

$qrystmt=$mysqli->prepare($query);
// You only need to call bind_param once
$qrystmt->bind_param("s",$submittedCourseId);
// get result and assign variables (prefix with db)

$qrystmt->execute(); 

$qrystmt->bind_result($dbCourseId,$dbModuleId,$dbCourseName,$dbModuleName);

$qrystmt->store_result();

$num = $qrystmt->num_rows();

if($num ==0){
echo "<p>Sorry, No Course was found with this Course ID '$course'</p>";
} else { 

$dataArray = array();

while ( $qrystmt->fetch() ) { 
// data array
$dataArray[$dbCourseId]['CourseName'] = $dbCourseName; 
$dataArray[$dbCourseId]['Modules'][$dbModuleId]['ModuleName'] = $dbModuleName; 
// session data
$_SESSION['idcourse'] = $dbCourseId;
$_SESSION['namecourse'] = $dbCourseName;

}

foreach ($dataArray as $foundCourse => $courseData) {

$output = ""; 

$output .= "<p><strong>Course:</strong> " . $foundCourse .  " - "  . $courseData['CourseName'] . "</p>";

$moduleHTML = ""; 
$moduleHTML .= '<select name="module" id="modulesDrop">'.PHP_EOL;
$moduleHTML .= '<option value="">Please Select</option>'.PHP_EOL;      
foreach ($courseData['Modules'] as $moduleId => $moduleData) {        

$moduleHTML .= "<option value='$moduleId'>" . $moduleId . " - " . $moduleData['ModuleName'] ."</option>".PHP_EOL;        
} 
}
$moduleHTML .= '</select>';

echo $output;

更新：

下面是视图页面源显示的内容：

        <form action="/u0000000/Mobile_app/create_session.php" method="post">
        <table>
        <tr>
        <th>Course: <select name="courses" id="coursesDrop">
<option value="">Please Select</option>
<option value='INFO101'>INFO101 - Bsc Information Communication Technology</option>
<option value='INFO102'>INFO102 - Bsc Computing</option>
</select></form><input id="courseSubmit" type="submit" value="Submit" name="submit" /></th>
        </tr>
        </table>
        </form>

        <p>Sorry, No Course was found with this Course ID 'INFO102'</p>  

Answer 1

尝试交换： foreach ($dataArray as $course => $courseData) 为了 foreach ($dataArray as $foundCourse => $courseData)

然后foreach 中的所有$courses 也应该重命名。您在页面顶部使用变量$course，即使它不能解决问题，它也会使代码更易于理解。

您还应该考虑使用 PDO 将您的数据库提供程序抽象出来 http://www.php.net/manual/en/faq.databases.php#faq.databases.mysql.deprecated

编辑：

代码说“这里是所有课程，现在如果你提交使用 $course 并获取它的详细信息”

但是您永远不会得到发布的课程，所以它会（碰巧）使用与脚本顶部的foreach“剩余”相同的命名变量。

我建议按照我上面说的那样做 if(isset($_POST['submit'])) 添加 $submittedCourseId = $_POST['coursesDrop'];

然后将 THAT 添加到您的绑定中 $qrystmt->bind_param("s",$submittedCourseId);

我认为这应该解决问题。