sql - Oracle SQL：使用过滤器获取第一行

Question

如何让第一行过滤掉一些值？我使用 row_number() over(partition by Name ORDER BY Date) 来获取订单（参见下面的示例）。但我需要，排名将从最后一次出现 Type = B 开始（参见预期输出）

SELECT Name, Age, Type, Date, 
       row_number() over(partition by Name ORDER BY Date) as Rank
FROM TableA;

例如 ：

Name   Age   Type   Date          Rank
Ben    12      A    2013/02/01    1 
Rod    14      A    2013/02/05    2
Zed    13      B    2013/03/09    3
Ken    12      A    2013/04/02    4 
Jed    14      B    2013/05/01    5
Mar    13      A    2013/05/04    6
Nic    12      A    2013/06/02    7
Jen    15      A    2013/06/09    8

预期输出：

Name   Age    Type   Date         Rank
Mar    13      A    2013/05/04    1
Nic    12      A    2013/06/02    2
Jen    15      A    2013/06/09    3

Answer 1

尝试

WITH qry AS 
(
  SELECT "Name", "Age", "Type", "Date", 
         ROW_NUMBER() OVER (PARTITION BY "Type" ORDER BY "Date") rank
  FROM TableA
)
SELECT "Name", "Age", "Type", "Date"
  FROM qry
 WHERE rank = 1

输出：

| NAME | AGE | TYPE |                            DATE |
-------------------------------------------------------
|  Ben |  12 |    A | February, 01 2013 00:00:00+0000 |
|  Zed |  13 |    B |    March, 09 2013 00:00:00+0000 |

这是SQLFiddle演示

Answer 2

还有另一种可能性：您可以将其包装在子查询中：

select
    t.*
  from
    (SELECT "Name", "Age", "Type", "Date", 
         ROW_NUMBER() OVER (PARTITION BY "Type" ORDER BY "Date") rank
       FROM TableA
     ) t
  where
    rank = 1

“子查询”和“公用表表达式( with)”的行为不同，因此请查看这些方法。这取决于您的情况，如果一个是强制性的（处理副作用）或者一个更快。在Oracle的情况下，甚至有一个materialize提示。“根据 SQL 标准，CTE 提供了优化围栏功能”（这是来自此处的改编引文）。

Answer 3

这也应该满足您的要求：

select t.*,
       row_number() over(partition by "Name" ORDER BY "Date") as Rank
from  TableA t
where "Date" >= (select max("Date") from TableA where "Type" = 'B')

Answer 4

尝试这个：

SELECT Name, Age, Type, Date, 
       row_number() over(partition by Name ORDER BY Date) as Rank
FROM TableA
WHERE type <> 'B'
AND Date > (
    SELECT max(Date)
    FROM TableA
    WHERE type = 'B'
);

请注意，这假设在与最后一个 B 完全相同的日期/时间没有 A。