SQL ORDER BY 两个具有优先级的字段，而不是一个接一个

Question

用于留言板系统。有一般帖子，以及对这些帖子的直接回复。

SELECT * FROM msgs ORDER BY msg_id ASC, msg_replyto ASC

不会给我想要的结果，因为它列出了所有的一般帖子，然后是所有的回复。我不知道该怎么表达我的要求，所以我所能做的就是给你模型表和数据。

这是示例表结构和行：

 msg_id | msg_text                               | msg_replyto
----------------------------------------------------------------
   1    | This is a post ID #1                   | NULL
   2    | This is a post ID #2                   | NULL
   3    | This is a direct reply to post ID #1   | 1
   4    | This is a post ID #4                   | NULL
   5    | This is a post ID #5                   | NULL
   6    | This is a direct reply to post ID #4   | 4
   7    | This is a post ID #7                   | NULL
   8    | This is a direct reply to post ID #3,  | 3
        | which is a direct reply to post ID #1. |

我尝试返回行的正确顺序如下：

 msg_id | msg_replyto
-----------------------
   1    | NULL
   3    | 1
   8    | 3
   2    | NULL
   4    | NULL
   6    | 4
   5    | NULL
   7    | NULL

其中，首先 msg_id 具有优先级 - 除非一行中的字段 msg_replyto 等于另一行的 msg_id，然后 msg_replyto 具有次要优先级。如果这没有意义，我很抱歉，我没有其他方法可以解释它。我试着搜索这个，但什至不知道如何措辞。

这是创建示例的实际 SQL。

CREATE TABLE IF NOT EXISTS `msgs` (
  `msg_id` int(11) NOT NULL AUTO_INCREMENT,
  `msg_text` varchar(200) NOT NULL,
  `msg_replyto` int(11) DEFAULT NULL,
  PRIMARY KEY (`msg_id`)
) ENGINE=MyISAM AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

INSERT INTO `msgs` (`msg_id`, `msg_text`, `msg_replyto`) VALUES
(1, 'This is a post ID #1', NULL),
(2, 'This is a post ID #2', NULL),
(3, 'This is a direct reply to post ID #1', 1),
(4, 'This is a post ID #4', NULL),
(5, 'This is a post ID #5', NULL),
(6, 'This is a direct reply to post ID #4', 4),
(7, 'This is a post ID #7 ', NULL),
(8, 'This is a direct reply to post ID #3, which is a direct reply to post ID #1', 3);
COMMIT;

Answer 1

作为一个普遍问题，没有简单的方法可以解决这个问题。你有一个图表行走问题。

但是有办法。您可以将每个方法的路径构造为字符串，然后按该字符串排序：

with recursive cte as (
      select msg_id, cast(lpad(msg_id, 5, '0') as char(10000)) as path
      from msgs
      where msg_replyto is null
      union all
      select m.msg_id, concat_ws('->', path, lpad(m.msg_id, 5, '0')) 
      from cte join
           msgs m
           on m.msg_replyto = cte.msg_id
     )
select m.*
from msgs m left join
     cte
     on m.msg_id = cte.msg_id
order by cte.path;

Here 是一个 dbfiddle。

Answer 2

下面应该得到你正在寻找的东西。它使用两个自连接来连接 msg_replyto 值，然后按 msg_id 和 msg_replyto 排序

SELECT C.*
FROM msgs  AS A
LEFT JOIN msgs  AS B
ON A.msg_id = IFNULL(B.msg_replyto,B.msg_id)
JOIN msgs  AS C
ON B.msg_id = IFNULL(C.msg_replyto,C.msg_id)
ORDER BY A.msg_id 
        ,B.msg_id

结果

msg_id	msg_text	msg_replyto
1	This is a post ID #1	NULL
3	This is a direct reply to post ID #1	1
8	This is a direct reply to post ID #3, which is a direct reply to post ID #1	3
2	This is a post ID #2	NULL
4	This is a post ID #4	NULL
6	This is a direct reply to post ID #4	4
5	This is a post ID #5	NULL
7	This is a post ID #7	NULL