Question

  with sample_data as (select '26.03.2015 14:10' as adate, 4 as type, 40 as object, 111 as barcode from dual union all
                       select '26.03.2015 14:09' as adate, 1 as type, 55 as object, 222 as barcode from dual union all
                       select '26.03.2015 14:08' as adate, 2 as type, 33 as object, 777 as barcode from dual union all
                       select '26.03.2015 14:08' as adate, 2 as type, 34 as object, null as barcode from dual union all
                       select '26.03.2015 13:20' as adate, 3 as type, 41 as object, null as barcode from dual union all
                       select '26.03.2015 12:00' as adate, 1 as type, 56 as object, 444 as barcode from dual union all
                       select '26.03.2015 11:59' as adate, 2 as type, 37 as object, 555 as barcode from dual union all
                       select '26.03.2015 11:59' as adate, 2 as type, 48 as object, null as barcode from dual)
  select 
  adate, type, object, barcode
  from sample_data 
  where type in (1, 2);

类型1已完成构建，类型2是程序集的一部分

我需要下一个结果

55 | 222 | 26.03.2015 14:08 | 33
55 | 222 | 26.03.2015 14:08 | 34
56 | 444 | 26.03.2015 11:59 | 37
56 | 444 | 26.03.2015 11:59 | 38

所以我们看到哪个对象 55 包含子对象 33 ， 34 和两个日期= 26.03.2015 14:08 和对象 56 包含子对象 37 ， 38 和两个日期= 26.03.2015 11:59

一开始我知道两个参数 - 完成建筑的日期和条形码，例如我的数据 - 日期 26.03.2015 14:09 和条形码 222 或日期< strong> 26.03.2015 12:00 和条形码 444

Answer 1

此查询完成此任务。子查询T1将“类型1”与“类型2”连接起来。 T2使用函数row_number（）枚举它们。上次查询仅过滤rn = 1的行。

with t1 as (
  select t.*, prior object pob, prior barcode pbc, 
      prior adate pad, level, sys_connect_by_path(object, '-') pth
    from sample_data t where level=2 
    connect by prior adate > adate and prior type = 1 and type = 2),
t2 as (select t1.*, 
    row_number() over (partition by object order by pad) rn from t1 )
select pob, pbc, adate, object from t2 where rn = 1

结果：

       POB        PBC ADATE               OBJECT
---------- ---------- ------------------- ----------
        55        222 2015-03-26 14:08:00         33
        55        222 2015-03-26 14:08:00         34
        56        444 2015-03-26 11:59:00         37
        56        444 2015-03-26 11:59:00         48

Answer 2

您必须定义组，然后才能将项目分配给组

with sample_data as (select '26.03.2015 14:10' as adate, 4 as type, 40 as object, 111 as barcode from dual union all
                     select '26.03.2015 14:09' as adate, 1 as type, 55 as object, 222 as barcode from dual union all
                     select '26.03.2015 14:08' as adate, 2 as type, 33 as object, 777 as barcode from dual union all
                     select '26.03.2015 14:08' as adate, 2 as type, 34 as object, null as barcode from dual union all
                     select '26.03.2015 13:20' as adate, 3 as type, 41 as object, null as barcode from dual union all
                     select '26.03.2015 12:00' as adate, 1 as type, 56 as object, 444 as barcode from dual union all
                     select '26.03.2015 11:59' as adate, 2 as type, 37 as object, 555 as barcode from dual union all
                     select '26.03.2015 11:59' as adate, 2 as type, 48 as object, null as barcode from dual),
     sd_groups as (select adate, lag(adate, 1) over (order by adate) adate_last, object, barcode from (
                     select * from sample_data where type = 1
                     union all select min(adate), null, null, null from sample_data))
select sg.object, sg.barcode, sd.adate, sd.object
from sample_data sd, sd_groups sg
where sd.type = 2 and sd.adate between sg.adate_last and sg.adate;

获取层次结构中对象的子对象

2 个答案: