Question

我有：

TABLE MESSAGES
 message_id | conversation_id | from_user | timestamp  |  message

我想：

1. SELECT * WHERE from_user <> id 
2. GROUP BY conversation_id
3. SELECT in every group row with MAX(timestamp) **(if there are two same timestamps in a group use second factor as highest message_id)** !!!
4. then results SORT BY timestamp

得到结果：

2|145|xxx|10000|message

6|1743|yyy|999|message

7|14|bbb|899|message

已淘汰

1|145|xxx|10000|message    <- has same timestamp(10000) as message(2) belongs to the same conversation(145) but message id is lowest  

5|1743|me|1200|message <- has message_from == me

具有相同时间戳的示例组

我希望从第3组开始，但我从查询

获得第2行

SELECT max(message_timestamp), message_id, message_text, message_conversationId
FROM MESSAGES
WHERE message_from <> 'me'
GROUP BY message_conversationId
ORDER by message_Timestamp DESC

我想从message_id＆amp;＆amp;时间戳，然后得到最大???

Answer 1

您的查询是基于GROUP BY的非标准使用（我认为SQLite仅允许与MySQL兼容）并且我完全不确定它是否会始终产生确定的结果。

另外，它在连接列上使用MAX()。除非你以某种方式确保两个（连接的）列具有固定的宽度，否则结果也不准确。

我会写这样的查询：

SELECT 
    m.message_timestamp, 
    m.message_id, 
    m.message_text,
    m.message_conversationId
FROM 
    ( SELECT message_conversationId         -- for every conversation
      FROM messages as m
      WHERE message_from <> 'me'            
      GROUP BY message_conversationId
    ) AS mc
  JOIN 
    messages AS m                           -- join to the messages
      ON  m.message_id =        
          ( SELECT mi.message_id            -- and find one message id
            FROM messages AS mi
            WHERE mi.message_conversationId      -- for that conversation
                  = mc.message_conversationId
              AND mi.message_from <> 'me'
            ORDER BY mi.message_timestamp DESC,  -- according to the
                     mi.message_id DESC          -- specified order
            LIMIT 1                              -- (this is the one part)
          ) ;

Answer 2

好吧，这比我想象的更简单：

基本上改变选择：

max(message_timestamp)

为：

max(message_timestamp || message_id)   
or  max(message_timestamp + message_id)

所以它会在时间戳的连接上搜索最多和message_id

PS。挖掘之后 - 只有当消息ID随时间戳增长（插入顺序被保留）时才能正常工作

编辑：

edit2：

为什么它有效呢？

SELECT max(message_timestamp+message_id), message_timestamp, message_id, message_conversationId, message_from,message_text
FROM MESSAGES
WHERE message_conversationId = 1521521
AND message_from <> 'me'
ORDER by message_Timestamp DESC

Answer 3

尝试使用以下sql按组实现两次目的。

select m.*
from
Messages m
-- 3. and then joining to get wanted output columns
inner join
(
    --2. then selecting from this max timestamp - and removing duplicates
    select conversation_id, max(timestamp), message_id
    from
    (
        -- 1. first select max message_id in remainings after the removal of duplicates from mix of cv_id & timestamp
        select conversation_id, timestamp, max(message_id) message_id
        from Messages
        where message <> 'me'
        group by conversation_id, timestamp
    ) max_mid
    group by conversation_id
) max_mid_ts on max_mid_ts.message_id = m.message_id
order by m.message_id;

http://goo.gl/MyZjyU

sqlite3按max查询，按第二因子过滤

3 个答案: