Question

    | RecordId | high_speed |   speed  | DateFrom   |  DateTo     |
    ---------------------------------------------------------------
    | 666542   |   60       |   10     | 09/11/2011 |  10/11/2011 |
    | 666986   |   20       |   20     | 11/11/2011 |  11/11/2011 |
    | 666996   |   0        |   0      | 13/11/2011 |  17/11/2011 |
    | 755485   |   0        |   0      | 01/11/2011 |  14/11/2011 |
    | 758545   |   70       |   50     | 15/11/2011 |  26/11/2011 |
    | 796956   |   40       |   40     | 09/11/2011 |  09/11/2011 |
    | 799656   |   25       |   20     | 09/11/2011 | 09/11/2011  |
    | 808845   |   0        |   0      | 15/11/2011 | 15/11/2011  |
    | 823323   |   0        |   0      | 15/11/2011 | 16/11/2011  |
    | 823669   |   0        |   0      | 17/11/2011 | 18/11/2011  |
    | 899555   |   0        |   0      | 18/11/2011 | 19/11/2011  |
    | 990990   |   20       |   10     | 12/11/2011 | 12/11/2011  |

这里，我想构建数据库视图，它组合了速度= 0的连续行。在这种情况下，DateFrom将是第一行和第一行的DateFrom值。 DateTo将是最后一行的DateTo值。表格结果如下：

| high_speed |    speed  | DateFrom    |    DateTo    |
---------------------------------------------------
|  60        |     10    |  09/11/2011 |  10/11/2011  |
|  20        |     20    |  11/11/2011 |  11/11/2011  |
|  0         |     0     |  13/11/2011 |  14/11/2011  |
|  70        |     50    |  15/11/2011 |  26/11/2011  |
|  40        |     40    |  09/11/2011 |  09/11/2011  |
|  25        |     20    |  09/11/2011 |  09/11/2011  |
|  0         |     0     |  15/11/2011 |  19/11/2011  |
|  20        |     10    |  12/11/2011 |  12/11/2011  |

有没有办法在数据库视图或函数中获得结果？

注意 - 1.删除了devID列。这是非常令人困惑的，而不是它增加了另一列来理解这个问题。另外，我需要添加一个“ Period ”列，即与“DateFrom”＆amp; “DateTo”栏目。

Answer 1

此查询使用分析函数lag()，lead()和一些带case ... when的逻辑提供所需的输出：

select high_speed, speed, datefrom, dateto, dateto-datefrom period
  from (
    select recordid, high_speed, speed, datefrom, 
      case when tmp = 2 then lead(dateto) over (order by recordid) 
                        else dateto end dateto, tmp 
      from (
        select test.*, case when speed <> 0 then 1 
                       when lag(speed) over (order by recordid) <> 0 then 2
                       when lead(speed) over (order by recordid) <> 0 then 3 
                       end tmp
          from test )
      where tmp is not null)
   where tmp in (1, 2) order by recordid

^SQLFiddle

Answer 2

这是同一解决方案的另一种方法。它使用lag()，lead()和partition by

与前一解决方案的不同之处在于此查询将行绑定连续的时间间隔，即：

考虑速度为0的所有行。

第1行：01/11至14/11
第2行：15/11至18/11
第3行：20/11至22/11

结果：

第1行：01/11至18/11（合并第1行和第2行）
第2行：20/11至22/11（第3行由于间隙19/11而分开）

另请注意，与15 / 11-15 / 11和13 / 11-17 / 11共享相同日期的句点将会破坏此查询。提供的示例数据具有此类句点。< / p>

-- for better understanding, start reading from the most nested query to the outer

-- QUERY 4: Removes duplicates

-- this query removes duplicates, because both border-rows on a multiple-row period will be identical
-- after the query 3

select distinct
    high_speed,
    speed,
    datefrom,
    dateto,
    dateto-datefrom period
from
(

-- QUERY 3: Selects border-rows and builds datefrom and dateto.

-- this query selects all border-rows, which have the datefrom and dateto data that we need
-- to build the bigger period row.
--
-- this query also builds the bigger period datefrom and dateto

select
    high_speed,
    speed,
    CASE WHEN is_previous_a_border = 0 and is_next_a_border = 1 then lag(datefrom) over (partition by speed order by datefrom)
         WHEN is_previous_a_border = 1 and is_next_a_border = 0 then datefrom
         WHEN is_previous_a_border = 1 and is_next_a_border = 1 then datefrom
         ELSE null END datefrom,
    CASE WHEN is_previous_a_border = 0 and is_next_a_border = 1 then dateto
         WHEN is_previous_a_border = 1 and is_next_a_border = 0 then lead(dateto) over (partition by speed order by datefrom)
         WHEN is_previous_a_border = 1 and is_next_a_border = 1 then dateto
         ELSE null END dateto
from (

-- QUERY 2: Create syntax-sugar

-- this query creates some syntax-sugar properties:
--    - "is_previous_a_border": defines if previous row is a border
--    - "is_next_a_border": defines if previous row is a border

select 
    high_speed,
    speed,
    datefrom,
    dateto,
    is_border,
    nvl(lag(is_border) over (partition by speed order by datefrom), 1) as is_previous_a_border,
    nvl(lead(is_border) over (partition by speed order by datefrom), 1) as is_next_a_border
 from (

-- QUERY 1: Create "is_border" property

-- this query creates the "is_border" property, which defines if a row is a border of a bigger multiple-row period
-- we use partition by to group rows and lag/lead to flag rows with consecutive periods
--
-- note that both border-rows of a bigger multiple-row period will have is_border = 1, while all rows in between
-- them, will have is_border = 0.

select
    high_speed,
    speed,
    datefrom,
    dateto,
    case when lead(datefrom) over (partition by speed order by datefrom) between datefrom and dateto + interval '1' day
         and lag(dateto) over (partition by speed order by datefrom) between datefrom - interval '1' day and dateto then 0
    else 1 end is_border
from
    test))
where is_border = 1)
order by
  speed, datefrom;

SQL Fiddle (with altered sample data)

SQL Oracle：组合连续的行

2 个答案: