sql - 选择时间范围外的不同行

Question

我正在尝试使用 SQL 根据一个条目与下一个条目之间的时间差来选择不同的数据条目。用一个例子更容易解释:

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

我想返回所有行，只要有一个限制，即如果有多个条目具有相同的“零件”编号，我只想检索相差至少 5 分钟的条目。

查询应该返回:

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

我使用的代码如下:

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可以建议替代方法？？

谢谢

Answer 1

救援分析函数。

您可以使用分析函数 LEAD 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，您就可以创建一个内联 View ，并只需选择下一个日期比当前日期晚 5 分钟以上的那些行。

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