android - sqlite3按最大值查询并按第二个因素过滤

Question

我有:

TABLE MESSAGES
 message_id | conversation_id | from_user | timestamp  |  message

我要:

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

与淘汰

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

我想从这个组中获取第 3 行，但我从查询中得到第 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 和 timestamp 做 union 然后得到 max 是什么意思？？？

Answer 1

您的查询基于 GROUP BY 的非标准使用(我认为 SQLite 允许这样做只是为了与 MySQL 兼容)，我完全不确定它是否会产生确定的结果时间。

此外，它在连接的列上使用 MAX()。除非您以某种方式确保两个(连接的)列具有固定宽度，否则结果也不会因此而准确。

我会这样写查询:

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)
          ) ;