如何在MySQL中的某些条件下为对字符串生成唯一的id

Question

我是SQL的新手，请帮我解决问题。

我有一个网站的网站私人消息，我在引擎之间迁移后得到（phpbb到drupal）

可以说，它有3列：

• mid - 用于消息ID
• thread - 用于线程id
• 收件人 - 用户ID，谁将收到邮件。

每条消息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。

问题是消息的线程 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

该表看起来像：

 ________________________________
| 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       |
|________|___________|___________|

我的目标是为每组配对行提供相同的线程ID，并使用同一对收件人：

 ________________________________
| 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       |
|________|___________|___________|

更新：

对于同一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。我想要的只是聚集线程中每2个用户之间存在大量现有消息）。

我想像某种循环一样，它将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：

• 1月中旬：（100,101）
• 2月中旬：（100,101）
• 3月中旬：（100,102）

然后给出相同的数组，这将是我想要的线程ID：

• （100,101）= 1
• （100,101）= 1
• （100,102）= 2

不确定，如果我的算法只能用SQL查询

Answer 1

应该从您的应用程序更好地处理这种更正，因为它比在这个功能样式Sql更容易处理逻辑。

这是完成它的一种方法（ 这假设所有相关（线程）对话一个接一个地进行，如您已确认 ）：

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 。 在这里测试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

Answer 2

我把它放到SQL小提琴中： http ：//sqlfiddle.com/#！2/83bdc/25

假设你的表名是messages那么这就是我要做的：

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......