Question

我正在尝试使用SQL根据一个条目和下一个条目之间的时间差选择不同的数据条目。用一个例子来解释更容易：

我的数据表有

Part    DateTime   
123     12:00:00
123     12:00:05
123     12:00:06
456     12:10:23
789     12:12:13
123     12:14:32

我希望返回所有行，但有一个限制，即如果有多个条目具有相同的“Part”号，我只想检索那些差异至少为5分钟的那些。

查询应返回：

Part    DateTime   
123     12:00:00
456     12:10:23
789     12:12:13
123     12:14:32

我正在使用的代码如下：

SELECT data1.*, to_char(data1.scan_time, 'yyyymmdd hh24:mi:ss') 

FROM data data1 

where exists
(
    select * 

    from data data2

    where data1.part_serial_number = data2.part_serial_number AND 
    data2.scan_time + 5/1440 >= data1.scan_time 
    and data2.info is null
)

order by to_char(data1.scan_time, 'yyyymmdd hh24:mi:ss'), data1.part_serial_number

不幸的是，这不起作用。有谁知道我做错了什么或者可以提出另一种方法吗？

由于

Answer 1

救援的分析功能。

您可以使用分析函数LEAD来获取零件下一行的数据。

SQL> ed
Wrote file afiedt.buf

  1  with x as (
  2    select 123 part, timestamp '2011-12-08 00:00:00' ts
  3      from dual
  4    union all
  5    select 123, timestamp '2011-12-08 00:00:05'
  6      from dual
  7    union all
  8    select 123, timestamp '2011-12-08 00:00:06'
  9      from dual
 10    union all
 11    select 456, timestamp '2011-12-08 00:10:23'
 12      from dual
 13    union all
 14    select 789, timestamp '2011-12-08 00:12:13'
 15      from dual
 16    union all
 17    select 123, timestamp '2011-12-08 00:14:32'
 18      from dual
 19  )
 20  select part,
 21         ts,
 22         lead(ts) over (partition by part order by ts) next_ts
 23*   from x
SQL> /

      PART TS                              NEXT_TS
---------- ------------------------------- -------------------------------
       123 08-DEC-11 12.00.00.000000000 AM 08-DEC-11 12.00.05.000000000 AM
       123 08-DEC-11 12.00.05.000000000 AM 08-DEC-11 12.00.06.000000000 AM
       123 08-DEC-11 12.00.06.000000000 AM 08-DEC-11 12.14.32.000000000 AM
       123 08-DEC-11 12.14.32.000000000 AM
       456 08-DEC-11 12.10.23.000000000 AM
       789 08-DEC-11 12.12.13.000000000 AM

6 rows selected.

完成后，您可以创建一个内联视图，只需选择下一个日期超过当前日期后5分钟的那些行。

SQL> ed
Wrote file afiedt.buf

  1  with x as (
  2    select 123 part, timestamp '2011-12-08 00:00:00' ts
  3      from dual
  4    union all
  5    select 123, timestamp '2011-12-08 00:00:05'
  6      from dual
  7    union all
  8    select 123, timestamp '2011-12-08 00:00:06'
  9      from dual
 10    union all
 11    select 456, timestamp '2011-12-08 00:10:23'
 12      from dual
 13    union all
 14    select 789, timestamp '2011-12-08 00:12:13'
 15      from dual
 16    union all
 17    select 123, timestamp '2011-12-08 00:14:32'
 18      from dual
 19  )
 20  select part,
 21         ts
 22    from (
 23      select part,
 24             ts,
 25             lead(ts) over (partition by part order by ts) next_ts
 26        from x )
 27   where next_ts is null
 28*     or next_ts > ts + interval '5' minute
SQL> /

      PART TS
---------- -------------------------------
       123 08-DEC-11 12.00.06.000000000 AM
       123 08-DEC-11 12.14.32.000000000 AM
       456 08-DEC-11 12.10.23.000000000 AM
       789 08-DEC-11 12.12.13.000000000 AM

Answer 2

AFJ，

让我们假设我们有一个新字段告诉我们在过去5分钟内是否存在此Part的previus条目，然后，将此字段设置为False的行我们得到结果。

select  
  Part, 
  DateTime,
  coalesce(
    (select distinct 1
     from data ds
     where ds.Part = d.Part
       and ds.DateTime between d.DateTime and d.DateTime - 5/1440
    )
    , 0) as exists_previous 
from data d

子查询检查它们是否是前5分钟inteval中具有相同Part的行

结果必须是：

Part    DateTime   exists_previous
123     12:00:00   0
123     12:00:05   1
123     12:00:06   1
456     12:10:23   0
789     12:12:13   0
123     12:14:32   0

现在，过滤以仅获取0：

的行

   select Part, DateTime from 
   (select  
      Part, 
      DateTime,
      coalesce(
        (select distinct 1
         from data ds
         where ds.Part = d.Part
           and ds.DateTime between d.DateTime and d.DateTime - 5/1440
        )
        , 0) as exists_previous 
    from data D
   ) T where T.exists_previous = 0

免责声明：未经测试。

Answer 3

这尚未得到验证，但基本上，诀窍是按部分和时间分组5分钟（无线）。

select part, min(scan_time)
from data
group by part, floor(scan_time/(5/1440))
order by scan_time;

选择时间范围之外的不同行

3 个答案: