1

我有以下数据

表格1

id   col1    col2       col3
----------------------------------
1    abc   01/01/2012    -
1    abc   01/01/2012    A
2    abc   01/01/2012    -
2    abc   01/02/2012    -
3    abc   01/02/2012    -
3    xyz   01/01/2012    -
4    abc   01/02/2012    -
4    xyz   01/01/2012    -
4    xyz   01/02/2012    -

以下是评估的顺序 -

if(col1 is false) then evaluate col2 if(col2 is false) then col3:

Col1 - xyz has first preference from all values in this column
col2 - min date
col3 - not '-' or min(col3)

我只想为每个 id 返回一行，如果 col1 失败转到 col2，如果失败则转到 col3 条件。从上表结果应该是

 id   col1    col2       col3
----------------------------------
 1    abc     01/01/2012  A
 2    abc     01/01/2012  -
 3    xyz     01/01/2012  -
 4    xyz     01/01/2012  -

我尝试使用密集等级，但没有帮助。我不确定如何使用任何可用的函数或 sql 逻辑来执行此逻辑。

for col1 - if more than one row for same code or xyz code then fail
for col2 - if more than one row with same min date then fail 
           [use this only if col1 condition fails]
4

2 回答 2

8

您可以在分析函数中指定许多要排序的条件

SELECT *
  FROM (SELECT id,
               col1,
               col2,
               col3,
               dense_rank() over (partition by id
                                      order by (case when col1 = 'xyz' 
                                                     then 1 
                                                     else 0 
                                                 end) desc,
                                               col2 asc,
                                               col3 asc) rnk
          FROM your_table)
 WHERE rnk = 1

我假设您希望dense_rank使用该dense_rank标签。你没有谈论你想如何处理关系或者关系是否可能，所以从问题本身并不清楚你是否要使用rank、dense_rank或row_number分析函数。如果您只获取每个 的最高排名行id，rank并且dense_rank行为相同，并且如果第一名存在平局，则将返回多行。 row_number将始终通过任意打破平局返回单行。如果您想获取 per 第一行以外的行id，那么您需要考虑 tie 并且您会从rankand获得不同的行为dense_rank。如果两行并列第一，dense_rank将分配第三行rnk2 而rank将其分配rnk为 3。

这似乎适用于您发布的示例数据

SQL> ed
Wrote file afiedt.buf

  1  with x as (
  2  select 1 id, 'abc' col1, to_date('01/01/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  3  select 1 id, 'abc' col1, to_date('01/01/2012', 'MM/DD/YYYY') col2, 'A' col3 from dual union all
  4  select 2 id, 'abc' col1, to_date('01/01/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  5  select 2 id, 'abc' col1, to_date('01/02/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  6  select 3 id, 'abc' col1, to_date('01/02/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  7  select 3 id, 'xyz' col1, to_date('01/01/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  8  select 4 id, 'abc' col1, to_date('01/02/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
  9  select 4 id, 'xyz' col1, to_date('01/01/2012', 'MM/DD/YYYY') col2, null col3 from dual union all
 10  select 4 id, 'xyz' col1, to_date('01/02/2012', 'MM/DD/YYYY') col2, null col3 from dual
 11  )
 12  SELECT *
 13    FROM (SELECT id,
 14                 col1,
 15                 col2,
 16                 col3,
 17                 dense_rank() over (partition by id
 18                                        order by (case when col1 = 'xyz'
 19                                                       then 1
 20                                                       else 0
 21                                                   end) desc,
 22                                                 col2 asc,
 23                                                 col3 asc) rnk
 24            FROM x)
 25*  WHERE rnk = 1
SQL> /

        ID COL COL2      C        RNK
---------- --- --------- - ----------
         1 abc 01-JAN-12 A          1
         2 abc 01-JAN-12            1
         3 xyz 01-JAN-12            1
         4 xyz 01-JAN-12            1
于 2012-07-12T17:57:08.680 回答
0 
with tmp(id, col1, col2, col3, col1b, col3b) as
(select distinct id, col1, col2, col3,
        case when col1 = 'xyz' then '0' else '1' || col1 end,
        case when col3 = '-' then '1' else '0' || col3 end
from Table1)

select t1.id, t1.col1, t1.col2, t1.col3 
from tmp t1
left join tmp t2 on t1.id = t2.id
                and t1.col1b > t2.col1b
left join tmp t3 on t1.id = t3.id
                and t1.col1b = t3.col1b
                and t1.col2 > t3.col2
left join tmp t4 on t1.id = t4.id
                and t1.col1b = t4.col1b
                and t1.col2 = t4.col2
                and t1.col3b > t4.col3b
where t2.id is null
  and t3.id is null
  and t4.id is null
于 2012-07-12T18:40:09.567 回答