如何计算Oracle SQL中的更改

Question

我有带有以下各列的下表：

HID_1 HID_2 Attr1 Attr2 Attr3 Attr4  Attr5    
123   111    wo     e    ak    ERR   20180630    
123   111    wo     e    ak    ERR   20180730     
123   111    wo     e    ak    ERR   20180830     
123   111    qe     e    ak    ERR   20180930    
123   111    qe     e    ak    ERR   20181030    
123   111    aa     a    ak    ERR   20181130

其中HID_1和HID_2是哈希ID广告，另外4列由group by语句定义，最后一个是time_id（每月的最后一天的日期）。 总的来说，在此表中，我有许多具有许多不同HID的记录。

我想将HID_2的许多更改（在Attr1-Attr4中）作为单独的列。 根据第一个示例，答案应该是这样的：

HID_1 HID_2 Attr1 Attr2 Attr3 Attr4  Attr5     Attr6    
123   111    wo     e    ak    ERR   20180630   0    
123   111    wo     e    ak    ERR   20180730   0    
123   111    wo     e    ak    ERR   20180830   0    
123   111    qe     e    ak    ERR   20180930   1     
123   111    qe     e    ak    ERR   20181030   0    
123   111    aa     a    ak    ERR   20181130   2

如何在Oracle sql数据库中执行操作？

Answer 1

尝试这个：

select t.* 
, case when attr1 != LAG(attr1, 1, attr1) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr2 != LAG(attr2, 1, attr2) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr3 != LAG(attr3, 1, attr3) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr4 != LAG(attr4, 1, attr4) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end as attr6
from t

Answer 2

我想你要：

select t.*,
       dense_rank() over (partition by hid_1, hid_2 order by min_attr5) as attr6
from (select t.*,
             min(attr5) over (partition by hid_1, hid_2, , attr1, attr2, attr3, attr4, seqnum_2 - seqnum) as min_attr5
      from (select t.*,
                   row_number() over (partition by hid_1, hid_2 order by attr5) as seqnum,
                   row_number() over (partition by hid_1, hid_2, attr1, attr2, attr3, attr4 order by attr5) as seqnum_2
            from t
      ) t;