[英]How to display correct options in drop down menu AND keep option displayed
[英]it does not display correct option details from drop down menu
我正在嘗試從“課程”下拉模塊中檢索所選課程的詳細信息,然后顯示一個模塊下拉菜單,其中列出了屬於該課程的所有模塊。
我遇到的問題是,我的課程下拉菜單中有以下兩個選項:
INFO101 - Business
INFO102 - ICT
由於某些奇怪的原因,每次我從下拉菜單中選擇頂部選項(INFO101)並單擊Submit按鈕時,它總是顯示其他課程詳細信息(INFO102),從而顯示屬於該課程的模塊,而不顯示其他課程課程。
我的問題是當我從下拉菜單中提交(INFO101)選項時,為什么它顯示其他課程的信息?
下面是mysqli代碼
$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 $output;
更新:
以下是視圖頁面源顯示的內容:
<form action="/u0000000/Mobile_app/create_session.php" method="post">
<table>
<tr>
<th>Course: <select name="courses" id="coursesDrop">
<option value="">Please Select</option>
<option value='INFO101'>INFO101 - Bsc Information Communication Technology</option>
<option value='INFO102'>INFO102 - Bsc Computing</option>
</select></form><input id="courseSubmit" type="submit" value="Submit" name="submit" /></th>
</tr>
</table>
</form>
<p>Sorry, No Course was found with this Course ID 'INFO102'</p>
嘗試交換: foreach ($dataArray as $course => $courseData)
for foreach ($dataArray as $foundCourse => $courseData)
然后foreach中的所有$course
也應重命名。 您可以在頁面頂部使用$course
變量,即使它不能解決問題,也會使代碼更易於理解。
您還應該研究使用PDO將您的數據庫提供程序抽象為http://www.php.net/manual/en/faq.databases.php#faq.databases.mysql.deprecated
編輯:
代碼說:“這是所有的課程,現在,如果您要提交使用$ course,並獲取其詳細信息,”
但是您根本不會獲得發布的課程,因此(恰好發生)將使用腳本頂部foreach
中“剩余”的相同命名變量。
我建議按照上面的說明做,然后在if(isset($_POST['submit']))
添加$submittedCourseId = $_POST['coursesDrop'];
然后在您的綁定$qrystmt->bind_param("s",$submittedCourseId);
添加THAT $qrystmt->bind_param("s",$submittedCourseId);
我認為應該解決問題。
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