我在我的dropdownmenu中有一個名爲$ moduleHTML的變量,$ moduleHTML輸出課程中的所有模塊。例如。課程:ICT,模塊:Web編程,現代數據庫應用,電子商務。如何正確地將變量放在下拉菜單中
現在,當我輸出$ moduleHTML時,它完美地工作,因爲它輸出課程中的所有模塊。但是,如果我嘗試將$ moduleHTML放在下拉菜單中,它只會在下拉菜單(即Web編程)中輸出一個模塊作爲選項。我想如何實現它,以便在下拉菜單中將多個模塊顯示爲多個選項?
下面是代碼:
<?php
$query = "SELECT cm.CourseId, cm.ModuleId,
c.CourseName,
m.ModuleName
FROM Course c
INNER JOIN Course_Module cm ON c.CourseId = cm.CourseId
JOIN Module m ON cm.ModuleId = m.ModuleId
WHERE
(c.CourseId = '".mysql_real_escape_string($courseid)."')
ORDER BY c.CourseName, m.ModuleId";
$num = mysql_num_rows($result = mysql_query($query));
$dataArray = array();
while ($row = mysql_fetch_array($result)) {
$dataArray[$row['CourseId']]['CourseName'] = $row['CourseName'];
$dataArray[$row['CourseId']]['Modules'][$row['ModuleId']]['ModuleName'] = $row['ModuleName'];
}
foreach ($dataArray as $courseId => $courseData) {
$output = "";
$output .= "<p><strong>Course:</strong> " . $courseId . " - " . $courseData['CourseName'] . "</p>";
foreach ($courseData['Modules'] as $moduleId => $moduleData) {
// elaborate module data
$moduleHTML = "";
$moduleHTML .= "<p>" . $moduleId . " - " . $moduleData['ModuleName'] ."</p>";
$output .= $moduleHTML;
}
}
echo $output;
?>
<form action="create_session.php" method="post" name="sessionform">
<table>
<tr>
<th>7: Module:</th>
<td><select name="module" class="modulesDrop">
<option value=""><?php echo $moduleHTML; ?></option>
</tr>
</table>
</form>
儘管這裏的所有答案都是一樣的,但我已經明白了這一點,因爲我最清楚的是恕我直言。 :) – Herbert