将选定的mysql_fetch_array（）值传递给处理脚本

Question

此脚本从users表中获取所有值，并在我正在构建的社交网络的“发送好友请求”类型场景中输出它们。那么如何成功将$ row ['id']传递给process-request.php？

$userid = $_SESSION['userid'];
$results = mysql_query("SELECT * FROM users");
while($row = mysql_fetch_array($results)) {
    if($userid != $row['user_pid']) {
    echo $row['firstname'] . " " . $row['lastname'];
    echo "<form method='POST' action='processing/process-request.php'>";
    echo '<input name="accepted" type="submit" value="Send User Request" /><br />';
    echo '<input name="AddedMessage" placeholder="Add a message?" type="textbox" />';
    echo '<br>Select Friend Type: ' . '<br />Full: ';
    echo '<input name="full_friend" type="checkbox"';
    echo '<input type="hidden" name="id" value="' . $row["id"] . '" />';
    echo '</form>';
    echo "<br /><hr />";
    } elseif ($userid == $row['user_pid']) {
    echo $row['firstname'] . " " . $row['lastname'];
    echo "<br />";
    echo "You all are already friends";
    }
}

Answer 1

由于您已经在使用会话，

$_SESSION['row-id'] = $row['id'];