我正在 PHP mysqli 中創(chuàng)建消息對話腳本。我有兩個表格收件箱和發(fā)送箱，這兩個表格相同的列我想加入這兩個表格。我想獲得兩個用戶之間的最后一條消息。收件箱表id  from_id     to_id    msg               sent_date1   2           3        hi how are you?   2019-12-05 04:14:202   3           2        fine              2019-12-05 05:15:583   2           3        hi                2019-12-05 03:20:344   5           2        hi                2019-12-05 08:30:40發(fā)件箱表id  from_id     to_id    msg               sent_date1   2           3        hi how are you?   2019-12-05 04:14:202   3           2        fine              2019-12-05 05:15:583   2           3        hi                2019-12-05 03:20:344   5           2        hi                2019-12-05 08:30:40這是我的源代碼<?phpif (isset($_SESSION['userid'])) {    $session_id = $_SESSION['userid'];}$sql = "SELECT *,    (SELECT username FROM users WHERE userid=from_id) AS from_username,    (SELECT username FROM users WHERE userid=to_id) AS to_username,    (SELECT username FROM users WHERE userid=?) AS my_username,    (SELECT profile_pic FROM users WHERE userid=from_id) AS from_profile_pic,    (SELECT profile_pic FROM users WHERE userid=to_id) AS to_profile_pic,    (SELECT profile_pic FROM users WHERE userid=?) AS my_profile_pic    FROM inbox WHERE from_id = ? OR to_id = ? ORDER BY id DESC";if ($stmt = $con->prepare($sql)) {    $stmt->bind_param('iiii', $session_id, $session_id, $session_id, $session_id);    $stmt->execute();}