我想知道是否有人可以幫忙?我試圖在用戶點擊一個鏈接時將2個從數據庫中提取的變量發送到另一個頁面。目前我只能發送一個。我知道我在下面做的是錯誤的.....基本上我想發送uninum和groupid到另一個頁面。當用戶點擊鏈接時如何將2個變量發送到不同的頁面
for ($i = 0; $i < $count; $i++){
$q = "SELECT participants.sname, participants.uninum, groups.groupid FROM participants INNER JOIN groups ON participants.uninum =
groups.uninum WHERE groups.groupid ='".$groups[$i]."'";
$result = mysqli_query ($dbcon, $q); // Run the query.
if ($result) { // If it ran, display the records.
// Table header.
echo '<table>
<tr><td><b>Edit</b></td>
<td><b>Surnname</b></td>
<td><b>University ID</b></td>
<td><b>Group</b></td>
</tr>';
// Fetch and display the records:
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo '<tr>
<td><a href="edit_group_member.php?uninum=' . $row['uninum'] . ' ?groupid=' . $row['groupid'] . ' ">Edit</a></td>
<td>' . $row['sname'] . '</td>
<td>' . $row['uninum'] . '</td>
<td>' . $row['groupid'] . '</td>
</tr>';
}
echo '</table>'; // Close the table.
mysqli_free_result ($result); // Free up the resources.
echo "<br><br>";
} else { // If it did not run OK.
// Public message:
echo '<p class="error">The current users could not be retrieved. We apologize for any inconvenience.</p>';
// Debugging message:
echo '<p>' . mysqli_error($dbcon) . '<br><br>Query: ' . $q . '</p>';
}
}
使用及到位秒? –
它需要'&groupid ='而不是'?groupid =' – Ankh