检索与输入的使用名称匹配的所有用户的上一条消息

Question

我想检索发送或接收的最后一条消息，该消息与通过AJAX在文本框中输入的用户名相匹配，但我无法编写正确的MySQL查询。

这是我的代码：

    $name=$_GET['name']; 
    $query_get_user_id = "SELECT id from login WHERE email = '".$_SESSION['login_user']."'"; 
    $result_get_userid=$mysqli->query($query_get_user_id);
    $user_row = mysqli_fetch_array($result_get_userid);
    $user_id=$user_row['id'];

    $query_second_user_id = "SELECT id from login WHERE name LIKE '".$name."%'"; 
    $result_second_get_userid=$mysqli->query($query_second_user_id);
    $user_second_row = mysqli_fetch_array($result_second_get_userid);
    $second_user_id=$user_second_row['id'];

    $query_get_all_users ="SELECT DISTINCT sender_id,receiver_id from message WHERE ((sender_id='".$user_id."' AND receiver_id='".$second_user_id."') OR (sender_id='".$user_id."' AND receiver_id='".$second_user_id."')) order by timestamp desc";

   $result_get_all_users=$mysqli->query($query_get_all_users);
   while($retrieve_user_row = mysqli_fetch_array($result_get_all_users)){  

这是我的表结构：

Answer 1

您可以使用变量根据时间戳对每个用户的消息进行排名，将最新消息排名为1，并仅保留排名为1的消息。

select * from (
    select m.*,
      @rn := if(u.id = @prevUserId, @rn+1, 1) rn,
      @prevUserId := u.id
    from users u 
    join messages m on u.id in (m.sender_id,m.receiver_id)
    where u.name like 'name%'
    order by u.id, timestamp desc
) t1 where rn = 1

Answer 2

抱歉，我弄错了，我只为登录用户或一个用户开发了查询，现在我明白了。

请尝试更新此查询。我在本地数据库上进行了测试，并按预期工作。

    SELECT *
FROM message mm
INNER JOIN
    (SELECT sender, MAX(createdtime) AS MaxDateTime
    FROM message
    WHERE sender IN 
    (select id from user where name like 'search_filter%')
    GROUP BY sender) groupedmm
ON mm.sender = groupedmm.sender 
AND mm.createdtime = groupedmm.MaxDateTime;