0

我正在努力制作一个GROUP BY连续的块，我使用以下两个作为参考：
- GROUP BY 用于 SQL 中的连续行
-我如何在 MySQL 中进行连续的分组？
- https://gcbenison.wordpress.com/2011/09/26/queries-that-group-tables-by-contiguous-blocks/

我试图用给定状态的开始和结束日期来封装期间的主要想法。与其他示例不同的是，我使用每个 room_id 的日期作为索引字段（而不是顺序 id）。

我的桌子：

room_id | calendar_date | state

样本数据：

1 | 2016-03-01 | 'a'
1 | 2016-03-02 | 'a'
1 | 2016-03-03 | 'a'
1 | 2016-03-04 | 'b'
1 | 2016-03-05 | 'b'
1 | 2016-03-06 | 'c'
1 | 2016-03-07 | 'c'
1 | 2016-03-08 | 'c'
1 | 2016-03-09 | 'c'
2 | 2016-04-01 | 'b'
2 | 2016-04-02 | 'a'
2 | 2016-04-03 | 'a'
2 | 2016-04-04 | 'a'

目标：

room_id | date_start | date_end   | state
1       | 2016-03-01 | 2016-03-03 | a
1       | 2016-03-04 | 2016-03-05 | b
1       | 2016-03-06 | 2016-03-09 | c
2       | 2016-04-01 | 2016-04-01 | b
2       | 2016-04-02 | 2016-04-04 | c

我在这方面做了两次尝试：
1）

SELECT
  rooms.row_new,
  rooms.state_new,
  MIN(rooms.room_id) AS room_id,
  MIN(rooms.state) AS state,
  MIN(rooms.date) AS date_start,
  MAX(rooms.date) AS date_end,
FROM
  (
    SELECT @r := @r + (@state != state) AS row_new,
      @state := state AS state_new,
      rooms.*
      FROM (
        SELECT @r := 0,
          @state := ''
      ) AS vars,
        rooms_vw
    ORDER BY room_id, date
  ) AS rooms
  WHERE room_id = 1
GROUP BY row_new
ORDER BY room_id, date
;

这非常接近工作，但是当我打印出row_new它开始跳转 (1, 2, 3, 5, 7, ...)

2)

SELECT 
    MIN(rooms_final.calendar_date) AS date_start,
    MAX(rooms_final.calendar_date) AS date_end,
    rooms_final.state,
    rooms_final.room_id,
    COUNT(*)
 FROM (SELECT 
     rooms.date,
     rooms.state,
     rooms.room_id,
     CASE
         WHEN rooms_merge.state IS NULL OR rooms_merge.state != rooms.state THEN
                     @rownum := @rownum+1
         ELSE
                     @rownum
         END AS row_num
            FROM rooms
            JOIN (SELECT @rownum := 0) AS row
       LEFT JOIN (SELECT rooms.date + INTERVAL 1 DAY AS date,
                         rooms.state,
                          rooms.room_id
                    FROM rooms) AS rooms_merge ON rooms_merge.calendar_date = rooms.calendar_date AND rooms_merge.room_id = rooms.room_id
            ORDER BY rooms.room_id, rooms.calendar_date
          ) AS rooms_final
 GROUP BY rooms_final.state, rooms_final.row_num
 ORDER BY room_id, calendar_date;

出于某种原因，这会返回一些null room_id 的结果，并且通常不准确。

4

2 回答 2

1

使用变量有点棘手。我会去：

SELECT r.state_new, MIN(r.room_id) AS room_id, MIN(r.state) AS state,
       MIN(r.date) AS date_start, MAX(r.date) AS date_end
FROM (SELECT r.*,
             (@grp := if(@rs = concat_ws(':', room, state), @grp,
                         if(@rs := concat_ws(':', room, state), @grp + 1, @grp + 1)
                       )
             ) as grp
    FROM (SELECT r.* FROM rooms_vw r ORDER BY ORDER BY room_id, date
         ) r CROSS JOIN
         (SELECT @grp := 0, @rs := '') AS params    
   ) AS rooms
WHERE room_id = 1
GROUP BY room_id, grp
ORDER BY room_id, date;

笔记：

  • 在一个表达式中分配一个变量并在另一个表达式中使用它是不安全的。MySQL 不保证表达式的求值顺序。
  • 在较新版本的 MySQL 中，您需要ORDER BY在子查询中执行。
  • 在最新版本中，您可以使用row_number()，大大简化了计算。
于 2018-06-18T23:25:20.083 回答
0

感谢@Gordon Linoff 为我提供了获得此答案的见解：

SELECT
  MIN(room_id) AS room_id,
  MIN(state) AS state,
  MIN(date) AS date_start,
  MAX(date) AS date_end
FROM
  (
    SELECT
  @r := @r + IF(@state <> state OR @room_id <> room_id, 1, 0) AS row_new,
      @state := state AS state_new,
      @room_id := room_id AS room_id_new,
      tmp_rooms.*
      FROM (
        SELECT @r := 0,
          @room_id := 0,
          @state := ''
      ) AS vars,
        (SELECT * FROM rooms WHERE room_id IS NOT NULL ORDER BY room_id, date) tmp_rooms
  ) AS rooms
GROUP BY row_new
order by room_id, date
;
于 2018-06-19T00:27:19.810 回答