这种情况如何结合sql select查询？

Question

我正在用 php/mysql 编写一个聊天应用程序

我有 3 个表：user, room 和 room_participant 具有这些结构：

user: id, username
room: id, title
room_participant: room_id, user_id

现在我想获取所有房间的列表以及每个房间中所有参与者的列表。

到目前为止，我只是从房间表中选择所有房间并遍历所有房间并从每个条目中选择用户信息，这是非常低效的。

有没有办法将所有这些选择组合成一个选择查询？

Answer 1

不经过测试不确定，但试一试：

SELECT
   room.*, 
   user.* 
FROM room 
  JOIN room_participant ON room_id = room_participant.id
  JOIN user ON room_participant.user_id = user.id
ORDER BY room.id

要对房间进行重复数据删除，请使用GROUP_CONCAT()

更新 GROUP_CONCAT()修改为返回id|username

SELECT
   room.id, room.name
   GROUP_CONCAT(CONCAT(user.id,'|',user.username)) AS userlist 
FROM room 
  JOIN room_participant ON room_id = room_participant.id
  JOIN user ON room_participant.user_id = user.id
GROUP BY room.id, room.name
ORDER BY room.id

将GROUP_CONCAT生成的userlist作为id|name,id|name,id|name，你可以使用PHP explode()来分隔它们。

// Split the list on the commas
$users = explode(",", $userlist);


$final_users = array();
// Then split each on the `|`
foreach ($users as $user) {
  $split_user = explode("|", $user);

  // Append each as a new associative array to $final_users
  $final_users[] = array('id' => $split_user[0], 'username' => $split_user[1]);
}

// Now you have an array of users as id, username
var_dump($final_users);

Answer 2

您可能希望执行两个查询，然后在使用 MySQL 的任何内容中进行匹配，例如PHP。

这两个：

SELECT id, title
  FROM room;

SELECT rp.room_id, rp.user_id, u.username
  FROM room_participant AS rp
       INNER JOIN user as u ON rp.user_id = u.id;

或者这两个：

SELECT id, username
  FROM user;

SELECT rp.room_id, rp.user_id, r.title
  FROM room_participant AS rp
       INNER JOIN room as r ON rp.user_id = r.id;

哪两个查询有意义取决于您对信息的真正用途。

您可以更进一步，分别选择所有三个：

SELECT *
  FROM room;

SELECT *
  FROM user;

SELECT *
  FROM room_participant;

注意：最好说明列，而不是使用“*”，以防将来您对这些查询并不真正感兴趣的新列被添加到表中。

显然，您必须在使用 MySQL 的任何内容中匹配所有内容，例如php。您可以根据所选信息创建房间和用户列表，然后将它们与以下内容匹配：

// Use MySQL to populate $roomList from database, then do...
foreach ($roomList as $room)
{
    $id    = $room['id'];
    $title = $room['title'];

    $this->roomList[$id] = new Room($id, $title);
}

// Use MySQL to populate $userList from database, then do...
foreach ($userList as $user)
{
    $id       = $user['id'];
    $username = $user['username'];

    $this->userList[$id] = new User($id, $username);
}

// Use MySQL to populate $roomParticipantList from database, then do...
foreach ($roomParticipantList as $roomParticipant)
{
    $room = $this->roomList[$roomParticipant['room_id']];
    $user = $this->userList[$roomParticipant['user_id']];

    // You could do one/both of these, depending on requirements.
    $room->enterUser($user);
    $user->joinRoom($room);
}