我正在尝试使用PARTITION BY OVER按特定列“分组”行。我有点理解PARTITION的使用，但是我想按日期“阻止”分区。例如，如果我们有
|col1|col2 |
| A |01/JAN/2012|
| A |01/FEB/2012|
| B |01/MAR/2012|
| B |01/APR/2012|
| A |01/MAY/2012|
我希望按col1进行分区，但我希望最后一个A与前两个A“不同”，因为它的日期分隔为'B'行。
如果我使用;
SELECT ROW_NUMBER() OVER (PARTITION BY col1 ORDER BY col2) AS RNUM, a.*
FROM table1 a;
它会屈服;
|RNUM|col1|col2 |
| 1| A |01/JAN/2012|
| 2| A |01/FEB/2012|
| 3| A |01/MAY/2012|
| 1| B |01/MAR/2012|
| 2| B |01/APR/2012|
但我真正想要的是;
|RNUM|col1|col2 |
| 1| A |01/JAN/2012|
| 2| A |01/FEB/2012|
| 1| B |01/MAR/2012|
| 2| B |01/APR/2012|
| 1| A |01/MAY/2012|
这是否可以使用PARTITION BY OVER？目前我已经退回到使用游标来解析数据并分配一个组ID，这样我就可以分开两个'A'序列，但这很慢。
谢谢，
标记。
答案 0 :(得分：4)
这可以通过几个分析来实现：
select col1, col2, row_number() over (partition by grp order by col2) rnum
from (select col1, col2, max(grp) over(order by col2) grp
from (select col1, col2,
case
when lag(col1) over (order by col2) != col1
then
row_number() over (order by col2)
when row_number() over(order by col2) = 1
then
1
end grp
from data));
即：
首先获取col1按col2日期更改排序的边界：
SQL> select col1, col2,
2 case
3 when lag(col1) over (order by col2) != col1
4 then
5 row_number() over (order by col2)
6 when row_number() over(order by col2) = 1
7 then
8 1
9 end grp
10 from data;
C COL2 GRP
- --------- ----------
A 01-JAN-12 1
A 01-FEB-12
B 01-MAR-12 3
B 01-APR-12
A 01-MAY-12 5
然后我们可以填写这些空值：
SQL> select col1, col2, max(grp) over(order by col2) grp
2 from (select col1, col2,
3 case
4 when lag(col1) over (order by col2) != col1
5 then
6 row_number() over (order by col2)
7 when row_number() over(order by col2) = 1
8 then
9 1
10 end grp
11 from data);
C COL2 GRP
- --------- ----------
A 01-JAN-12 1
A 01-FEB-12 1
B 01-MAR-12 3
B 01-APR-12 3
A 01-MAY-12 5
然后通过按row_number()排序并在col2
grp
答案 1 :(得分：0)
首先，你应该找到每条记录的GROUP_ID，以便将所有类似的COL1排序到不同的组，如果它们之间有差距的话。然后在带有COL1的OVER语句中使用此GROUP_ID：
SELECT ROW_NUMBER() OVER (PARTITION BY Group_id,col1 ORDER BY col2) AS RNUM, a3.*
FROM
(
select a1.*,
(select count(*) from t a2 where
a2.col1<>a1.col1
AND
a2.col2<a1.col2) as GROUP_ID
from t a1
) a3
order by col2
答案 2 :(得分：0)
请参阅下面的方法，它类似于Dazzal的答案，有点不同的逻辑：
步骤1：
--find the swhitches to new groups
select col1, col2,
case when nvl(lag(col1) over (order by col2),sysdate) <> col1 then 1 end as new_grp
from data;
COL1 COL2 NEW_GRP
A January, 01 2012 1
A February, 01 2012 (null)
B March, 01 2012 1
B April, 01 2012 (null)
A May, 01 2012 1
步骤2：
--identify/mark the groups
select col1, col2, sum(new_grp) over (order by col2) as grp
from(
select col1, col2,
case when nvl(lag(col1) over (order by col2),sysdate) <> col1 then 1 end as new_grp
from data)
;
COL1 COL2 NEW_GRP
A January, 01 2012 1
A February, 01 2012 1
B March, 01 2012 2
B April, 01 2012 2
A May, 01 2012 3
步骤3：
--find the row_number within group
select col1, col2, row_number() over(partition by grp order by col2) rn
from(
select col1, col2, sum(new_grp) over (order by col2) as grp
from(
select col1, col2,
case when nvl(lag(col1) over (order by col2),sysdate) <> col1 then 1 end as new_grp
from data
)
);
COL1 COL2 NEW_GRP
A January, 01 2012 1
A February, 01 2012 2
B March, 01 2012 1
B April, 01 2012 2
A May, 01 2012 1
答案 3 :(得分：0)
您不需要分区。您需要将日期转换为DD / MM / YYYY格式并订购。或者，如果必须，那么你可以通过MM部分进行分区，它可以给你01,02,03 ...并且可以根据需要进行分区并轻松转换为数字。但是你不需要这一切......不要让你的查询复杂化。始终保持简单。外部查询仅用于将日期重新格式化为DD / MON / YYYY格式：
SELECT val, to_char(to_date(dt, 'DD/MM/YYYY'), 'DD/MON/YYYY') formatted_date
FROM
( -- Format your date to DD/MM/YYYY and order by it --
SELECT 'A' val, to_char(to_date('01/JAN/2012'), 'DD/MM/YYYY') dt FROM dual
UNION
SELECT 'A', to_char(to_date('01/FEB/2012'), 'DD/MM/YYYY') FROM dual
UNION
SELECT 'B',to_char(to_date('01/MAR/2012'), 'DD/MM/YYYY') FROM dual
UNION
SELECT 'B',to_char(to_date('01/APR/2012'), 'DD/MM/YYYY') FROM dual
UNION
SELECT 'A',to_char(to_date('01/MAY/2012'), 'DD/MM/YYYY') FROM dual
ORDER BY 2
)
/
您的日期按照您的要求订购：
VAL FORMATTED_DATE
-------------------
A 01/JAN/2012
A 01/FEB/2012
B 01/MAR/2012
B 01/APR/2012
A 01/MAY/2012