Oracle 按组划分为基于日期的序列

Question

我正在尝试使用 PARTITION BY OVER 按某些列“分组”行。我对 PARTITION 的使用有所了解，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|

我想按 col1 进行分区，但我希望最后一个 A 与前两个“不同”，因为它在日期方面被“B”行分隔。

如果我使用；

SELECT ROW_NUMBER() OVER (PARTITION BY col1 ORDER BY col2) AS RNUM, a.* 
FROM table1 a;

它会产生；

|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|

这可以使用 PARTITION BY OVER 吗？目前我已经退回到使用游标来解析数据并分配一个组 id，这样我就可以分隔两个“A”序列，但这很慢。

谢谢，

标记。

Answer 1

您不需要分区。您需要将日期转换为 DD/MM/YYYY 格式并订购它们。或者，如果必须，那么您可以按 MM 部分进行分区，这将为您提供 01,02,03... 并且可以根据需要进行分区并轻松转换为数字。但是您不需要所有这些……不要使您的查询复杂化。始终保持简单。外部查询只是将您的日期重新格式化为 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
)
/

您的日期按照您的意愿排序：

VAL FORMATTED_DATE
-------------------
A   01/JAN/2012
A   01/FEB/2012
B   01/MAR/2012
B   01/APR/2012
A   01/MAY/2012

Answer 2

请看下面我的方法，这与 Dazzal 的回答类似，逻辑略有不同：

SQL FIDDLE

第一步：

--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

第二步：

--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

第三步：

--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

Answer 3

首先，您应该为每条记录找到 GROUP_ID，以便将所有相似的 COL1 排序到不同的 GROUPS（如果它们之间有间隔）。然后在 OVER 语句中使用这个 GROUP_ID 和 COL1：

SQLFiddle demo

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

Answer 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 更改顺序的边界col2 日期：

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

然后是通过col2 排序并在grp 上进行分区来分配row_number() 的情况

小提琴：http://sqlfiddle.com/#!4/4818c/1