Question

我需要一个消息列表，其中每个消息都是当前用户与其他用户之间“对话”中的最新消息。

描述了相同的查询in this question

我到目前为止的代码是：

t1 = Arel::Table.new(:messages, :as => 't1')
t2 = Arel::Table.new(:messages, :as => 't2')

convs1 = t1.
          project(
                  t1[:receiver_user_id].as('other_user_id'), 
                  t1[:receiver_user_id].as('receiver_user_id'), 
                  t1[:sender_user_id].as('sender_user_id'), 
                  t1[:created_at].as('created_at')
                 ).
          where(t1[:sender_user_id].eq(user.id))

convs2 = t2.project(
                  t2[:sender_user_id].as('other_user_id'), 
                  t2[:receiver_user_id].as('receiver_user_id'), 
                  t2[:sender_user_id].as('sender_user_id'), 
                  t2[:created_at].as('created_at')
                 ).
          where(t2[:receiver_user_id].eq(user.id))

conv = convs1.union(convs2)

首先，我收到一个错误：

ActiveRecord::StatementInvalid: Mysql2::Error: You have an error in your SQL syntax; check \
  the manual that corresponds to your MySQL server version for the right syntax to use near \
  'UNION SELECT `t2`...

如果我在下面生成的sql中手动将“UNION”替换为“UNION ALL”，则此方法有效。上面代码中的conv.to_sql产生：

SELECT `t1`.`receiver_user_id` AS other_user_id, 
       `t1`.`receiver_user_id` AS receiver_user_id, `
        t1`.`sender_user_id` AS sender_user_id, 
       `t1`.`created_at` AS created_at 
  FROM `messages` `t1` 
 WHERE `t1`.`sender_user_id` = 50 
UNION 
SELECT `t2`.`sender_user_id` AS other_user_id, 
       `t2`.`receiver_user_id` AS receiver_user_id, 
       `t2`.`sender_user_id` AS sender_user_id, 
       `t2`.`created_at` AS created_at 
  FROM `messages` `t2` 
 WHERE `t2`.`receiver_user_id` = 50

知道为什么会出现MySQL UNION错误。它是一个arel bug吗？其次，非常感谢任何有关完成查询的帮助。

更新使用Arel :: Nodes :: Union.new工作

Answer 1

我认为这可能是一个mysql错误，这是一个mySQL错误文本。 here中讨论了类似的问题，但不完全是这个问题。

尝试迁移到另一个sql server，然后再次检查，或union all是否有效，然后使用this：

conv = convs1.union(convs2, :all)

基于文档。

Answer 2

问题实际上是sql中的括号。如果我跑，它会起作用：

Message.find_by_sql conv.to_sql.delete('()')

删除了前导＆＃34;（＆＃34;和尾随＆＃34;）＆＃34;

很奇怪..我不知道如何将其链接以完成查询。（Arel :: Nodes :: Union没有组方法）。这是Rails 3.1.4

Answer 3

我有类似的问题并按如下方式解决：

  def last_messages
    Message.find_by_sql("
          SELECT  messages.*,
                  (IF(recipient_id = #{id}, 0,1)) as outlast,
                  users.avatar_name,
                  users.name 
          FROM messages 
          INNER JOIN users 
              ON users.id=(IF(recipient_id = #{id}, sender_id,recipient_id))
          WHERE messages.id IN 
              ( SELECT max(id)
                FROM messages
                WHERE recipient_id = #{id} OR sender_id = #{id}
                GROUP BY (IF(recipient_id = #{id}, sender_id, recipient_id))
              )
          ORDER BY messages.id DESC")
  end

Answer 4

这是我在最后使用的代码

all_msgs = Message.where("messages.sender_user_id = ? OR messages.receiver_user_id = ?",
                                     user.id, user.id)

msg_ids = all_msgs.select("sender_user_id, receiver_user_id, max(id) as max_id")
.group(:sender_user_id, :receiver_user_id).map { |m| m.max_id }

all_msgs = all_msgs.where(:id => msg_ids)

arel union和来自消息的最新对话

4 个答案: