[英]Oracle sql query to group consecutive records by date
使用以下示例数据,我试图以相同的速率对记录进行分组。
id start_date end_date rate
-----------------------------------------------------------------
1 01/01/2017 12:00:00 am 01/01/2017 12:00:00 am 300
1 02/01/2017 12:00:00 am 02/01/2017 12:00:00 am 300
1 03/01/2017 12:00:00 am 03/01/2017 12:00:00 am 300
1 04/01/2017 12:00:00 am 04/01/2017 12:00:00 am 1000
1 05/01/2017 12:00:00 am 05/01/2017 12:00:00 am 500
1 06/01/2017 12:00:00 am 06/01/2017 12:00:00 am 500
1 07/01/2017 12:00:00 am 07/01/2017 12:00:00 am 1000
1 08/01/2017 12:00:00 am 08/01/2017 12:00:00 am 1000
1 09/01/2017 12:00:00 am 09/01/2017 12:00:00 am 300
我尝试过的
select distinct id, mn_date, mx_date,rate
from (
select id, min(start_date) over (partition by grp order by start_date) mn_date,
max(end_date) over(partition by grp order by start_date desc) mx_date, rate
from (
select t.*, row_number() over(partition by id order by start_date) -row_number() over(partition by rate order by start_date)grp
from t
)
)
order by mn_date;
输出:
id mn_date mx_date rate
--------------------------------------------------------
1 01/01/2017 12:00:00 am 03/01/2017 12:00:00 am 300
1 04/01/2017 12:00:00 am 04/01/2017 12:00:00 am 1000
1 05/01/2017 12:00:00 am 06/01/2017 12:00:00 am 500
1 07/01/2017 12:00:00 am 09/01/2017 12:00:00 am 300
1 07/01/2017 12:00:00 am 09/01/2017 12:00:00 am 1000
所需输出:
id mn_date mx_date rate
--------------------------------------------------------
1 01/01/2017 12:00:00 am 03/01/2017 12:00:00 am 300
1 04/01/2017 12:00:00 am 04/01/2017 12:00:00 am 1000
1 05/01/2017 12:00:00 am 06/01/2017 12:00:00 am 500
1 07/01/2017 12:00:00 am 08/01/2017 12:00:00 am 1000
1 09/01/2017 12:00:00 am 09/01/2017 12:00:00 am 300
最终结果按连续日期分组:(感谢Gordon)
select id, min(start_date), max(end_date), rate
from (
select id, start_date, end_date, rate, seqnum_i-seqnum_ir grp, sum(x) over(partition by id order by start_date) grp1
from (
select t.*,
row_number() over (partition by id order by start_date) as seqnum_i,
row_number() over (partition by id, rate order by start_date) as seqnum_ir,
case when LEAD(start_date) over (partition by id order by start_date)= end_date + 1
then 0
else 1
end x
from t
)
)
group by id, grp+grp1, rate
order by min(start_date);
假设我们可以只使用start_date
来识别相邻的记录(即没有空位),那么可以使用行数差方法:
select id, min(start_date) as mn_date, max(end_date) as mx_date, rate
from (select t.*,
row_number() over (partition by id order by start_date) as seqnum_i,
row_number() over (partition by id, rate order by start_date) as seqnum_ir
from t
) t
group by id (seqnum_i - seqnum_ir), rate;
若要查看其工作原理,请查看子查询的结果。 您应该能够“看到”两个行号的差值如何以相同的速率定义相邻记录的组。
我发现最后一个值未正确分组,因为X的计算未处理NULL返回,因此我将其更改为:
,CASE
WHEN LEAD (start_date)
OVER (PARTITION BY id ORDER BY start_date)
IS NULL
THEN
0
WHEN LEAD (start_date)
OVER (PARTITION BY id ORDER BY start_date) =
end_date + 1
THEN
0
ELSE
1
END
x
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