Oracle Analytic函数-重置窗口条款 [英] Oracle Analytic functions - resetting a windowing clause

问题描述

我有以下数据集.

create table t1 (
  dept number,
  date1 date
);

Table created.

insert into t1 values (100, '01-jan-2013');
insert into t1 values (100, '02-jan-2013');
insert into t1 values (200, '03-jan-2013');
insert into t1 values (100, '04-jan-2013');
commit;

我的目标是创建一个等级列，该等级列在每次部门变更时都会重置.我可以用于"partition by"子句的最接近的列是dept，但这不会给我想要的结果.

MY goal is to create a rank column that resets each time the department changes. The closest column that I can use for "partition by" clause is dept, but that won't give me the desired result.

SQL> select * from t1;

      DEPT DATE1
---------- ---------
       100 01-JAN-13
       100 02-JAN-13
       200 03-JAN-13
       100 04-JAN-13

select dept,  
       date1,
       rank () Over (partition by dept order by date1) rnk
from t1
order by date1;

      DEPT DATE1            RNK
---------- --------- ----------
       100 01-JAN-13          1
       100 02-JAN-13          2
       200 03-JAN-13          1
       100 04-JAN-13          3

所需的输出如下.最后一次rnk = 1是因为Jan-04记录是更改后的第一条记录.

The desired output is as follows. The last rnk=1 is becuase the Jan-04 record is the first record after the change.

      DEPT DATE1            RNK
---------- --------- ----------
       100 01-JAN-13          1
       100 02-JAN-13          2
       200 03-JAN-13          1
       100 04-JAN-13          1  <<<----------

有指针吗?

推荐答案

这有点复杂.代替使用rank()等，使用lag()查看何时发生了更改.然后对标志求和.

This is a little complicated. Instead of using rank() or the like, use lag() to see when something changes. Then do a cumulative sum of the flag.

select dept, date1,
       CASE WHEN StartFlag = 0 THEN 1
            ELSE 1+StartFlag+NVL(lag(StartFlag) over (order by date1),0)
       END as rnk
from (select t1.*,
             (case when dept = lag(dept) over (order by date1)
                   then 1
                   else 0
              end) as StartFlag
      from t1
     ) t1
order by date1;

此处是SQLFiddle.

Here is the SQLFiddle.

这是戈登在编辑我自己的答案.哎呀.原始查询的查询率为90％.它确定了编号应增加的组，但未在组内分配编号.我可以使用另一个row_number()级别来做到这一点，例如:

This is Gordon editing my own answer. Oops. The original query was 90% of the way there. It identified the groups where the numbers should increase, but did not assign the numbers within the groups. I would do this with another level of row_number() as in:

select dept, date1,
       row_number() over (partition by dept, grp order by date1) as rnk
from (select dept, date1, startflag,
             sum(StartFlag) over (partition by dept order by date1) as grp
      from (select t1.*,
                   (case when dept = lag(dept) over (order by date1)
                         then 0
                         else 1
                    end) as StartFlag
            from t1
           ) t1
     ) t1
order by date1;

因此，总体思路如下.首先使用lag()来确定组的开始位置(即，从一个日期到下一个日期的部门更改在哪里).然后，通过累加总和为这些分配一个组ID".这些是要枚举的记录.最后一步是使用row_number()枚举它们.

So, the overall idea is the following. First use lag() to determine where a group begins (that is, where there is a department change from one date to the next). Then, assign a "group id" to these, by doing a cumulative sum. These are the records that are to be enumerated. The final step is to enumerate them using row_number().

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