Question

不确定该问题的措词是否应该清晰（这是我能提出的最好的答案），但是这里有一个示例可以使事情变得清晰。我有一个Chats视图，该视图应该总结两个人之间的对话历史。该视图包括以下几列：Sender，Recipient，Timestamp，LatestMessage和UnreadMessageCount。

Chats视图的列均来自表Direct_Messages，该表存储有关系统用户之间交换的各个聊天消息的详细信息。以下是其列： ID，Sender，Recipient，Body，Timestamp，TimeRead（如果收件人未读取邮件，则为null）。视图的Timestamp和LatestMessage列具有两个参与者之间最新的直接消息的值（最新的Timestamp FWIW）。

问题的根源实际上是，在Sender视图中仅存在 Recipient，Chats个复合列的一个排列，即两位参与者之间的最新交流。例如，如果加里（Gary）向巴里（Barry）发送了“嗨”消息，那么巴里回答了“你好”（Hello），这两个家伙之间Chats中唯一的条目应该是Sender为“巴里”， Recipient为'Gary'，Timestamp为Barry的回复时间戳，LatestMessage为'Hello'，UnreadMessageCount为Recipient尚未收到的邮件数阅读。

我尝试使用GROUP BY "Sender", "Recipient" OR "Recipient", "Sender"，但它只返回两列：一列由Barry，Gary分组；另一列由Barry，Gary分组。另一个由加里·巴里（Gary，Barry）分组

这是我的代码：

SELECT Sender AS Sender,
       Recipient AS Recipient,
       Timestamp AS Timestamp,
       Body AS LatestMessage,
       (SUM(CASE WHEN TimeRead IS NULL THEN 1 ELSE 0 END) ) AS UnreadMessageCount
FROM Direct_Messages
GROUP BY Sender, Recipient OR Recipient, Sender
ORDER BY Timestamp DESC

编辑：这是Direct_Messages表中的示例数据，以及Chats视图中的相应输出

来自Direct_Messages

ID          Sender  Recipient   Body    Timestamp                   TimeRead
148567984   Gary    Barry       Hi      2018-12-12 23:53:39.487     2018-12-12 23:55:45
1668701120  Barry   Gary        Hello   2018-12-12 23:54:49.326     NULL

结果Chats：

Sender  Recipient   Timestamp                 LatestMessage UnreadMessageCount
Gary    Barry       2018-12-12 23:53:39.487   Hi            0
Barry   Gary        2018-12-12 23:54:49.326   Hello         1

Answer 1

您可以“预存储”您的数据，以使来自每个用户组合的消息始终处于同一方向。

样本，如果您的数据是：

Sender Recipient
A ---> B
B ---> A

您将其更改为：

U1     U2
B ---> A (changed)
B ---> A

像这样：

SELECT (case when Sender > Recipient then Sender else Recipient end) AS u1,
       (case when Sender > Recipient then Recipient else Sender end) AS u2,
       Timestamp AS Timestamp,
       Body AS LatestMessage,
       (SUM(CASE WHEN TimeRead IS NULL THEN 1 ELSE 0 END) ) AS UnreadMessageCount
FROM Direct_Messages_cooked
GROUP BY 
     (case when Sender > Recipient then Sender else Recipient end), 
     (case when Sender > Recipient then Recipient else Sender end) 
ORDER BY Timestamp DESC

注意：注意性能（我想这并不重要，因为您将问题标记为sqlite）

您可以使用CTE来预查询数据并获取更具可读性的查询：

with Direct_Messages_coocked as
(
    select
      (case when Sender > Recipient then Sender else Recipient end) AS U1,
      (case when Sender > Recipient then Recipient else Sender end) AS U2,
      *
    from Direct_Messages
)
SELECT U1 AS U1,
       U2 AS U2,
       Timestamp AS Timestamp,
       Body AS LatestMessage,
       (SUM(CASE WHEN TimeRead IS NULL THEN 1 ELSE 0 END) ) AS UnreadMessageCount
FROM Direct_Messages_coocked
GROUP BY U1, U2
ORDER BY Timestamp DESC

Answer 2

通过将MIN()和MAX()与多个参数一起使用，您可以获得所需的大部分内容。这些具有多个参数的 scalar 函数在其他数据库中的作用类似于LEAST()和GREATEST()：

SELECT MIN(Sender, Recipient) AS u1,
       MAX(Sender, Recipient) AS u2,
       MAX(Timestamp) AS Timestamp,
       -- Body AS LatestMessage,
       (COUNT(*) - COUNT(TimeRead)) as UnreadMessageCount
FROM Direct_Messages_cooked
GROUP BY u1, u2
ORDER BY MAX(Timestamp) DESC

挑战在于获取最新方法。您可以通过条件聚合和其他JOIN来获得此功能：

SELECT MIN(dmc.Sender, dmc.Recipient) AS u1,
       MAX(dmc.Sender, dmc.Recipient) AS u2,
       MAX(dmc.Timestamp) AS Timestamp,
       MAX(CASE WHEN dmc.Timestamp = dmc2.Timestamp THEN Body END) AS LatestMessage,
       (COUNT(*) - COUNT(dmc.TimeRead)) as UnreadMessageCount
FROM Direct_Messages_cooked dmc JOIN
     (SELECT MIN(Sender, Recipient) AS u1,
             MAX(Sender, Recipient) AS u2,
             MAX(Timestamp) AS Timestamp
      FROM Direct_Messages_cooked
      GROUP BY u1, u2
     ) dmc2
     ON dmc2.u1 = MIN(dmc.Sender, dmc.Recipient) AND
        dmc2.u2 = MAX(dmc.Sender, dmc.Recipient)
GROUP BY u1, u2
ORDER BY dmc2.Timestamp DESC

Answer 3

在@Gordon Linoff和@dani herrera的有见地的答案的基础上，我设法进行了调整，并提出了针对我特定问题的简洁解决方案，尽管在我最初的问题的更广泛的范围内，从我的观察来看，@ Gordon的答案似乎是，以更充分地解决该问题。这是我设法提出的：

SELECT Sender AS Sender,
       Recipient AS Recipient,
       Timestamp AS Timestamp,
       Body AS LatestMessage,
       (COUNT( * ) - COUNT(TimeRead) ) AS UnreadMessageCount
  FROM Direct_Messages
 GROUP BY (
              SELECT MAX(Sender, Recipient) 
          ),
          (
              SELECT MIN(Sender, Recipient) 
          )
 ORDER BY Timestamp DESC

如何通过两列组合键的排列对行进行分组

3 个答案: