好吧,我只是將它更改爲$ _POST,它現在正在工作。我不確定這是否是捷徑。至少現在工作。如果你想幫助我,你可以幫我縮小代碼。謝謝從下拉菜單中獲取相同ID的整行
<?php
$conn = new mysqli('localhost', 'root', 'jared17', 'hbadb')
or die ('Cannot connect to db');
$result = $conn->query("select * from english");
echo "<html>";
echo "<body>";
echo "<form method = POST>";
echo "<select name = 'Students'>";
while ($row = $result->fetch_assoc()) {
$LRN = $row['LRN'];
$Last = $row['Last_Name'];
$First = $row['First_Name'];
$Lvl = $row['Level'];
$Q1 = $row['Q1'];
$Q2 = $row['Q2'];
$Q3 = $row['Q3'];
$Q4 = $row['Q4'];
$Final = $row['FINAL'];
echo '<option value="'.$LRN.'|'.$Last.', '.$First.'|'.$Lvl.'|'.$Q1.'|'.$Q2.'|'.$Q3.'|'.$Q4.'|'. $Final.'">'.$Last.', '.$First.'</option>';
}
echo "</select>";
echo "<input type='submit' name='submit' value='Show'>";
echo "</form>";
$show = $_POST['Students'];
$show_explode = explode('|', $show);
echo "<table><tr><th>LRN</th><th>Name</th><th>Level</th><th>Q1</th><th>Q2</th><th>Q3</th><th>Q4</th><th>Final</th></tr>";
echo "<tr><td>". $show_explode[0]."</td><td>". $show_explode[1]."</td><td>". $show_explode[2]."</td><td>". $show_explode[3]."</td><td>". $show_explode[4]."</td><td>". $show_explode[5]."</td><td>". $show_explode[6]."</td><td>". $show_explode[7]."</td></tr>";
echo "</table>";
echo "</body>";
echo "</html>";
?>
在分配變量之前,不需要取消設置變量。 – Barmar
*我想要做的是獲得下拉菜單*要求或問題的相同ID的整行? – devpro
不要把所有的信息都放在'