我有下表,其中包含以下列:
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
如果HID_1和HID_2是哈希标识,则其他4列由group by语句定义,最后一列是time_id(该月的最后一天的日期)。通常在这个表中我有更多的记录,有很多不同的HID。
我希望将HID_2的一些更改(在Attr1 - Attr4中)作为单独的列进行计算。根据第一个例子,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
我怎么能在Oracle sql数据库中做?
试试这个:
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
我想你想要:
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;