我有一个如下表
row_wid id code sub_code item_nbr orc_cnt part_cnt variance reporting_date var_start_date
1 1 ABC PQR 23AB 0 1 1 11-10-2019 NULL
2 1 ABC PQR 23AB 0 1 1 12-10-2019 NULL
3 1 ABC PQR 23AB 1 1 0 13-10-2019 NULL
4 1 ABC PQR 23AB 1 2 1 14-10-2019 NULL
5 1 ABC PQR 23AB 1 3 2 15-10-2019 NULL
我只需要对id,code,sub_code和item_nbr的每个组合使用min(reporting_date)的min(reporting_date)更新var_start_date列,直到方差字段为零为止。 方差= 0的行应为空var_start_date。并且此后的下一行应具有下一个min(var_start_date。)。仅供参考,方差计算为par_cnt-orc_cnt
所以我的输出应该像这样-
row_wid id code sub_code item_nbr orc_cnt part_cnt variance reporting_date var_start_date
1 1 ABC PQR 23AB 0 1 1 11-10-2019 11-10-2019
2 1 ABC PQR 23AB 0 1 1 12-10-2019 11-10-2019
3 1 ABC PQR 23AB 1 1 0 13-10-2019 NULL
4 1 ABC PQR 23AB 1 2 1 14-10-2019 14-10-2019
5 1 ABC PQR 23AB 1 3 2 15-10-2019 14-10-2019
我正在尝试使用下面的查询编写一个函数,以将数据分为几组。
SELECT DISTINCT MIN(reporting_date)
OVER (partition by id, code,sub_code,item_nbr ORDER BY row_wid ),
RANK() OVER (partition by id, code,sub_code,item_nbr ORDER BY row_wid)
AS rnk,id, code,sub_code,item_nbr,orc_cnt,part_cnt,variance,row_wid
FROM TABLE T1
。但是不知道如何包括方差字段来拆分集合。
答案 0 :(得分:1)
我建议:
IF give_code NOT IN ( SELECT course_code FROM module WHERE course_code = ('WSD' OR 'DDM' OR 'NSF'))THEN
SIGNAL SQLSTATE '45000'SET MESSAGE_TEXT = 'Please Check Entered Course And Try Again'; END IF;
请注意,这不使用select t.*,
(case when variance <> 0
then min(reporting_date) over (partition by id, code, sub_code, item_nbr, grouping)
end) as new_reporting_date
from (select t.*,
sum(case when variance = 0 then 1 else 0 end) over (partition by id, code, sub_code, item_nbr) as grouping
from t
) t;
。它应该比答案更有效。
答案 1 :(得分:0)
尝试如下
SELECT T.*, CASE WHEN T.variance = 0 THEN NULL ELSE MIN(reporting_date) OVER (PARTITION BY T1.RANK ORDER BY T1.RANK) END AS New_var_start_date
FROM mytbl T
LEFT JOIN (
SELECT row_wid, variance, COUNT(CASE variance WHEN 0 THEN 1 END) OVER (ORDER BY row_wid ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) +1 AS [Rank]
FROM mytbl
) T1 ON T.row_wid = T1.row_wid