Question

我正在努力制作一个GROUP BY个连续的块，我使用了以下两个作为参考：
- GROUP BY for continuous rows in SQL
- How can I do a contiguous group by in MySQL?
- https://gcbenison.wordpress.com/2011/09/26/queries-that-group-tables-by-contiguous-blocks/

我尝试使用给定状态的开始和结束日期来封装句点的主要想法。与其他示例不同的是，我使用 date per room_id 作为索引字段（而不是序列ID）。

我的表：

room_id | calendar_date | state

示例数据：

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：
的 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 row_new 时，它开始跳跃（1,2,3,5,7，...）< / p>

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的结果以及通常不准确的结果。

Answer 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;

注意：

在一个表达式中分配变量并在另一个表达式中使用它是不安全的。 MySQL不保证表达式的评估顺序。
在更新版本的MySQL中，您需要在子查询中执行ORDER BY。
在最新版本中，您可以使用row_number()，大大简化了计算。

Answer 2

感谢@Gordon Linoff给我的见解以获得这个答案：

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
;

MySQL - 由Contigous Blocks组

2 个答案: