3个月组

Question

我在Oracle中创建了一个select，它返回机组人员每月过夜的次数(如果有的话)：

CRE_ALPHA  CRE_NAME  MONTH  YEAR  NIGHT_STOPS  
---------- --------- ------ ----- ------------
AAC        Adinda    6      2013  8  
AAC        Adinda    7      2013  9  
AAC        Adinda    8      2013  2  
AAC        Adinda    9      2013  7  
AAC        Adinda    10     2013  4  
CCU        Cristiano 6      2013  5  
CCU        Cristiano 7      2013  6  
CCU        Cristiano 8      2013  3  
CCU        Cristiano 9      2013  11
CVA        Carine    7      2013  9
CVA        Carine    9      2013  10
CVA        Carine    10     2013  10

现在，在3个月的基础上，有一个18晚停留的限制.因此，我想分组的任何三个月后，有超过18个晚上停止。结果应该是这样的：

CRE_ALPHA  CRE_NAME  TIMESPAN        NIGHT_STOPS
---------- --------- --------------- ------------
AAC        Adinda    6/2013-8/2013   19
AAC        Adinda    7/2013-9/2013   18
CCU        Cristiano 7/2013-9/2013   20
CVA        Carine    7/2013-9/2013   19
CVA        Carine    8/2013-10/2013  20

请注意，如果一个月的夜间停止时间为零，则没有行，但我希望得到3个月的结果，包括0的结果。

有人能帮忙吗？

如果可以帮助，请在下面进行完整选择：

ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MM-YYYY hh24:mi:ss';
SELECT cre_id, cre_alpha, cre_first_name, cre_last_name, Maand, Jaar, count(*) "Night stops"
FROM
  (SELECT cre_id, cre_alpha, cre_first_name, cre_last_name, pos_crb_iata_code, dst, det, dsa, (dst - Prev_end_time) * 1440 stop_over, EXTRACT(MONTH FROM dst) Maand, EXTRACT(YEAR FROM dst) Jaar
  FROM
    (SELECT cre_id, cre_alpha, cre_first_name, cre_last_name, pos_crb_iata_code, dst, dsa, det, dea, LAG(det) OVER (ORDER BY cre_alpha, dst) Prev_end_time
    FROM
      (SELECT cre_id, cre_alpha, cre_first_name, cre_last_name, pos_crb_iata_code,
      COALESCE(flt_mvt_db, flt_com_dep_blk, pog_std, gco_start, oth_std, rsv_std) as dst,
      COALESCE(flt_mvt_ab, flt_com_arr_blk, pog_sta, gco_end, oth_sta, rsv_sta) as det,
      COALESCE(flt_apt_iata_code_dep, pog_apt_iata_code_from, gco_apt_iata_code, rsv_apt_iata_code) as dsa,
      COALESCE(flt_apt_iata_code_arr, pog_apt_iata_code_to, gco_apt_iata_code, rsv_apt_iata_code) as dea
      FROM
        (SELECT DISTINCT cre_id, cre_alpha, cre_first_name, cre_last_name, pos_crb_iata_code
        FROM master.crews, master.assignments, master.positions
        WHERE asg_pos_id = pos_id AND asg_cre_id = cre_id AND asg_d_type <> 'LEA' 
        AND asg_start_time BETWEEN '01-JUN-2013' AND '01-NOV-2013'
        ORDER BY cre_alpha) tab1, master.assignments
      FULL OUTER JOIN master.flights ON master.assignments.asg_flt_id = master.flights.flt_id
      FULL OUTER JOIN master.positionings ON master.assignments.asg_pog_id = master.positionings.pog_id
      FULL OUTER JOIN master.ground_courses ON master.assignments.asg_gco_id = master.ground_courses.gco_id
      FULL OUTER JOIN master.other_duties ON master.assignments.asg_oth_id = master.other_duties.oth_id
      FULL OUTER JOIN master.reserves ON master.assignments.asg_rsv_id = master.reserves.rsv_id
      WHERE asg_d_type <> 'LEA' AND asg_d_type <> 'STP' AND asg_cre_id = tab1.cre_id
      AND asg_start_time BETWEEN '01-JUN-2013' AND '02-NOV-2013' AND asg_actif = 'Y'
      ORDER BY cre_alpha, asg_start_time)
    )
  WHERE pos_crb_iata_code <> dsa
  AND EXTRACT(DAY FROM dst) - EXTRACT(DAY FROM Prev_end_time) >= 1)
WHERE stop_over > 240
GROUP BY cre_id, cre_alpha, cre_first_name, cre_last_name, Maand, Jaar
ORDER BY cre_alpha;

Answer 1

你可以使用分析功能来实现你想要的。在当前查询的基础上，如下所示：

select *
from (
    select cre_alpha, cre_name,
    month month_end, year year_end,
    sum(night_stops) over (
        partition by cre_alpha, cre_name
        order by year * 12 + month
        range between 2 preceding and current row
    ) as night_stops
    from (
      ... your current query ...
    ) t
) m
where night_stops >= 18

注意：

• 查询返回3个月期间的结束(年份/月)。您还必须扩展它来打印句号的开始。
• 我使用了条件>= 18来匹配您的输出，尽管文本中说它是> 18。
• 窗口子句range between 2 preceding and current row和order子句year * 12 + month确保了三个月的窗口，而不仅仅是连续的三行。如果基础查询中缺少月份，则这与此相关。

玩得开心。