Question

我有一张包含以下信息的表格：

数据样本

**Table 1**  
palletNumber-- event-- --recordDate
-----1-----------A-------01/11/2015 01:00
-----1-----------B-------01/11/2015 02:00
-----1-----------C-------01/11/2015 03:00
-----1-----------D-------01/11/2015 04:00
-----2-----------A-------01/11/2015 01:10
-----2-----------C-------01/11/2015 01:15
-----2-----------E-------01/11/2015 01:20

我想通过palletNumber选择recordDate序列中表中出现的所有可能事件组合。我尝试使用行号，超过分区的各种语句，但这并没有让我接近我正在寻找的东西......任何方向去哪儿？

这将是输出表，例如：

**Table 2**  
event1-- event2--
---A------B------
---B------C------
---C------D------
---A------C------
---C------E------

谢谢，

Answer 1

您可以使用lag()或lead()

获取上一个或下一个活动

select event,
       lead(event) over (partition by palletnumber order by recorddate) as next_event
from datasample;

如果你想消除重复，我倾向于使用group by，因为这也可以计算每对出现的次数：

select event, next_event, count(*) as cnt
from (select event,
             lead(event) over (partition by palletnumber order by recorddate) as next_event
      from datasample
     ) ds
group by event, next_event;

Answer 2

用例：

select case when palletNumber = 1 then event else null end as event1,
       case when palletNumber = 2 then event else null end as event2,
       recordDate
  from table1

然后，您可以使用潜在客户 / 滞后或总和（）/ 分组来处理数据把它放在一排。

假设事件1/2每日只有一条记录

  select recordDate, max (event1), max (event2)
    from (  select case when palletNumber = 1 then event else null end as event1,
                   case when palletNumber = 2 then event else null end as event2,
                   recordDate
              from table1
          order by recordDate) tab2
group by recordDate

SQL Oracle - 根据日期查找可能的所有事件组合

2 个答案: