Question

我的Oracle数据库中有以下结构：

Date          Allocation  id
2015-01-01    Same        200
2015-01-02    Good        200
2015-01-03    Same        200
2015-01-04    Same        200
2015-01-05    Same        200
2015-01-06    Good        200

我希望有一个查询只需检查前几天的连续日，并获得分配为"Same"的计数。

我想按日期选择，例如2015-01-05 示例输出：对于日期2015-01-05，计数为3。

新问题。根据Lukas Eder的查询，计数始终为1或2。但预期的是3。从2015-01-03到2015-01-05。

Date          Allocation  id
2015-01-01    Same        400
2015-01-02    Good        400
2015-01-03    Same        400
2015-01-04    Same        400
2015-01-05    Same        400
2015-01-06    Good        400

Lukas Eder的代码

 SELECT c
    FROM (
      SELECT allocation, d, count(*) OVER (PARTITION BY allocation, part ORDER BY d) AS c
      FROM (
        SELECT allocation, d,
               d - row_number() OVER (PARTITION BY allocation ORDER BY d) AS part
        FROM t
      )
    )
    WHERE d = DATE '2015-01-05';

Answer 1

您可以使用不同的行来标识组。这就是你的查询正在做的事情。如果您想要特定日期的值，请使用：

select t.*
from (select t.*,
             (row_number() over (order by date) -
              row_number() over (partition by allocation order by date)
             ) as grp
      from table t
     ) t
where d = date '2015-01-05';

我最好猜测为什么问题中的版本不起作用是因为您的date可能有时间组件。我假设d是date列。

编辑：

要获取计数，请使用子查询和count(*)作为分析函数：

select t.*
from (select t.*, count(*) over (partition by grp, allocation) as cnt
      from (select t.*,
                   (row_number() over (order by date) -
                    row_number() over (partition by allocation order by date)
                   ) as grp
            from table t
           ) t
      ) t
where d = date '2015-01-05';

或者，如果您只想要计数，则可以使用聚合：

      select date, count(*) as cnt
      from (select t.*,
                   (row_number() over (order by date) -
                    row_number() over (partition by allocation order by date)
                   ) as grp
            from table t
           ) t
      group by date, allocation
      having date = date '2015-01-05';

Answer 2

给它一个旋转：

    create table mbtmp(d_date date, alloc varchar2(20), id number);

insert into mbtmp values (to_date('2015-01-01','yyyy-mm-dd'),'SAME',200);
insert into mbtmp values (to_date('2015-01-02','yyyy-mm-dd'),'GOOD',200);
insert into mbtmp values (to_date('2015-01-03','yyyy-mm-dd'),'SAME',200);
insert into mbtmp values (to_date('2015-01-04','yyyy-mm-dd'),'SAME',200);
insert into mbtmp values (to_date('2015-01-05','yyyy-mm-dd'),'SAME',200);
insert into mbtmp values (to_date('2015-01-06','yyyy-mm-dd'),'GOOD',200);
COMMIT;

SELECT max(length(sames) - length(replace(sames,'|',null)) - 1) from ( 
select regexp_substr( pth,'[|SAME]+') sames 
FROM(
   SELECT sys_connect_by_path(alloc,'|') pth  
    from mbtmp m
    WHERE d_Date > ( select max(d_date) from mbtmp 
                     where d_Date < to_date('2015-01-05','yyyy-mm-dd')
                      And alloc != 'SAME' )
    START WITH d_Date = to_date('2015-01-05','yyyy-mm-dd')
    CONNECT BY Prior d_date = d_date + 1));

现在，如果不是连续日规则，即 - 如果即使日期列表中存在间隙仍然可以计算，那么您可以使用更简单的查询：

select count(d_Date)
    FROM mbtmp
    WHERE d_Date > ( select max(d_date) from mbtmp 
                      where d_Date < to_date('2015-01-05','yyyy-mm-dd')
                      And alloc != 'SAME' )
    and   d_date <= to_date('2015-01-05','yyyy-mm-dd')

Oracle - 连续数天符合相同的给定条件

2 个答案: