我是SQL的新手，请帮我解决问题。
我有一个用于网站私人消息的表，我在引擎之间迁移后得到的（phpbb到drupal）
可以说，它有3列：
每封邮件都显示在收件人和邮箱的邮箱中，因此每封邮件都有2个字符串。
问题是消息的主题 ID不正确（基本上，它们等于 mid ），因此，2个用户之间的对应关系显示为一百个单独的线程。
例如：
user 100 wrote message to user 101
user 101 replied to user 100
user 102 wrote message to user 100
表格如下：
________________________________
| mid | thread | recipient |
| | | |
| 1 | 1 | 101 | ← message 1 in recipient's mailbox
| 1 | 1 | 100 | ← message 1 in author's mailbox
| 2 | 2 | 100 | ...
| 2 | 2 | 101 |
| 3 | 3 | 100 |
| 3 | 3 | 102 |
|________|___________|___________|
我的目标是为具有相同收件人对的每对配对行提供相同的线程ID：
________________________________
| mid | thread | recipient |
| | | |
| 1 | 1 | 101 | }
| 1 | 1 | 100 | } Correspondence between 100 and 101 must
| 2 | 1 | 100 | } have the same thread id (1)
| 2 | 1 | 101 | }
| 3 | 2 | 100 |
| 3 | 2 | 102 |
|________|___________|___________|
对于相同收件人对之间的所有当前对话，线程ID必须相同。 （不要担心进一步的对话，用户可以选择在编写消息时打开新线程，这将生成新的线程ID，或者在现有线程中回复，这将标记具有现有线程ID的消息。我想要的只是在线程中每2个用户之间收集大量现有消息）。
我想，就像某种循环一样，它会搜索每个中间的所有收件人id并在升序排序后将它们合并到数组中：
然后给出相同的数组，这将是我想要的线程ID：
不确定，如果我的算法只能用SQL查询
答案 0 :(得分：0)
这种更正应该从你的应用程序中更好地处理，因为它比那个功能样式的Sql更容易处理逻辑。
以下是完成此任务的一种方法（这假设所有相关（线程）对话都是一个接一个地进行，因为您已经确认）：
SELECT mid,
thread,
recipient
FROM (
SELECT row,
mid,
recipient,
IF(row % 2 = 0, @evenThread := thread, NULL) AS tempThread,
IF(row % 2 = 1, @evenThread, thread) AS thread
FROM (
SELECT @row := @row + 1 as row,
mid,
IF(@row = 1, @thread := thread, NULL) AS tempThread,
IF(@row % 2 = 1, IF(@row = 1, @rec1 := recipient, @rec1 := @rec3), NULL) AS tempOddRow,
IF(@row % 2 = 1, @rec3 := recipient, recipient) AS recipient,
IF(@row % 2 = 0 AND @row > 2, IF((recipient != @rec1 OR @rec3 != @rec2) AND (recipient != @rec2 OR @rec3 != @rec1), @isDifferent := true, @isDifferent := false), NULL) AS isDifferent,
IF(@row % 2 = 0, @rec2 := recipient, NULL) AS tempEvenRow,
IF(@row % 2 = 0, IF(@isDifferent, @thread := @thread + 1, @thread), @thread) AS thread
FROM (SELECT @row := 0) AS r,
(SELECT @thread := 0) as t,
(SELECT @isDifferent := false) as d,
(SELECT @rec1 := 0) AS r1,
(SELECT @rec2 := 0) AS r2,
(SELECT @rec3 := 0) AS r3,
messages --your table name
ORDER BY row DESC
) AS temp,
(SELECT @evenThread := 0) AS t
) AS corrected
ORDER BY row
这只是选择查询。在此测试http://sqlfiddle.com/#!9/61247/2。您可以将其复制到临时表中，并首先确认所有数据是否正确，而不是更新原始表，这将使调试更加困难。有点像：
CREATE TABLE messages_new LIKE messages; --new table name given
INSERT INTO messages_new (mid, thread, recipient)
SELECT mid,
thread,
recipient
FROM (
SELECT row,
mid,
recipient,
IF(row % 2 = 0, @evenThread := thread, NULL) AS tempThread,
IF(row % 2 = 1, @evenThread, thread) AS thread
FROM (
SELECT @row := @row + 1 as row,
mid,
IF(@row = 1, @thread := thread, NULL) AS tempThread,
IF(@row % 2 = 1, IF(@row = 1, @rec1 := recipient, @rec1 := @rec3), NULL) AS tempOddRow,
IF(@row % 2 = 1, @rec3 := recipient, recipient) AS recipient,
IF(@row % 2 = 0 AND @row > 2, IF((recipient != @rec1 OR @rec3 != @rec2) AND (recipient != @rec2 OR @rec3 != @rec1), @isDifferent := true, @isDifferent := false), NULL) AS isDifferent,
IF(@row % 2 = 0, @rec2 := recipient, NULL) AS tempEvenRow,
IF(@row % 2 = 0, IF(@isDifferent, @thread := @thread + 1, @thread), @thread) AS thread
FROM (SELECT @row := 0) AS r,
(SELECT @thread := 0) as t,
(SELECT @isDifferent := false) as d,
(SELECT @rec1 := 0) AS r1,
(SELECT @rec2 := 0) AS r2,
(SELECT @rec3 := 0) AS r3,
messages
ORDER BY row DESC
) AS temp,
(SELECT @evenThread := 0) AS t
) AS corrected
ORDER BY row;
DROP TABLE messages;
ALTER TABLE messages_new RENAME TO messages; --back to old name
答案 1 :(得分：0)
我把它放在一个SQL小提琴中：http://sqlfiddle.com/#!2/83bdc/25
假设您的表名是messages，那么我就是这样做的：
Select messages.*, FirstThread
From messages
Inner Join
(
-- Find a pair of dudes for each message and show the earliest thread of each
Select ThreadsForPairs.*, FirstThread From
(
Select mid
, Min(recipient) AS FirstDude
, Max(recipient) AS SecondDude
, thread
From messages
Group By mid, thread
) ThreadsForPairs
Left Join
(
-- Find the earliest thread for every pair of dudes
Select FirstDude, SecondDude, Min(thread) AS FirstThread
From
-- For every message, get the dudes involved
(
Select mid
, Min(recipient) AS FirstDude
, Max(recipient) AS SecondDude
, thread
From messages
Group By mid, thread
) PairsForMessages
Group By FirstDude, SecondDude
) FirstThreadForPairs
ON ThreadsForPairs.FirstDude = FirstThreadForPairs.FirstDude
AND ThreadsForPairs.SecondDude = FirstThreadForPairs.SecondDude
) FirstThreadForEveryMessage
On messages.mid = FirstThreadForEveryMessage.mid
正如您将看到的，我不完全匹配您的输出，因为消息ID 3将具有线程3而不是2，但我认为它是相同的想法......