Question

我很想解决一个容易解决的问题（至少对我来说是MySQL / SqlServer！）

我会简化问题。假设我有下表：

表格投票

ID  ID_IDEA DATE_VOTE   with ID_IDEA FK(IDEA.ID)
1   3       10/10/10
2   0       09/09/10
3   3       08/08/10
4   3       11/11/10
5   0       06/06/10
6   1       05/05/10

我正在尝试为每个单独的想法找到最新的投票，这意味着我只想返回ID为4,2和6的行。

在Oracle看来，如果不使用SUM（），AVG等函数，就无法使用GROUP BY。我对它应该如何工作感到有点困惑。

请指教，

感谢。

Answer 1

SELECT id,
       id_idea,
       date_vote
FROM   (SELECT id,
               id_idea,
               date_vote,
               Row_number() over (PARTITION BY id_idea 
                                      ORDER BY date_vote DESC NULLS LAST) AS rn
        FROM   VOTE) AS t
WHERE  rn = 1

Answer 2

据我了解，您正在寻找：

SELECT id_idea, max(date_vote)
FROM vote
GROUP BY id_idea

编辑：第二个想法，如果你需要获得完整的行：

SELECT v.*
FROM vote v
  JOIN (SELECT id_idea, max(date_vote) as max_date
        FROM vote
        GROUP BY id_idea) t
    ON t.id_idea = v.id_idea AND t.max_date = v.date_vote

Answer 3

如果您只需汇总：

，就不应该使用分析函数来处理这样的查询

SQL> create table vote(id,id_idea,date_vote)
  2  as
  3  select 1, 3, date '2010-10-10' from dual union all
  4  select 2, 0, date '2010-09-09' from dual union all
  5  select 3, 3, date '2010-08-08' from dual union all
  6  select 4, 3, date '2010-11-11' from dual union all
  7  select 5, 0, date '2010-06-06' from dual union all
  8  select 6, 1, date '2010-05-05' from dual
  9  /

Table created.

SQL> select max(id) keep (dense_rank last order by date_vote) id
  2       , id_idea
  3       , max(date_vote) date_vote
  4    from vote
  5   group by id_idea
  6  /

        ID    ID_IDEA DATE_VOTE
---------- ---------- -------------------
         2          0 09-09-2010 00:00:00
         6          1 05-05-2010 00:00:00
         4          3 11-11-2010 00:00:00

3 rows selected.

与分析变体相比：

1）它有效（如果你在'AS t'中删除'AS'，分析的也可以工作）

2）它更短

3）它更清晰（好吧，这是主观的）

4）它的性能稍微高一点，见：

这是聚合查询的计划：

Execution Plan
----------------------------------------------------------
Plan hash value: 2103353780

---------------------------------------------------------------------------
| Id  | Operation          | Name | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------
|   0 | SELECT STATEMENT   |      |     3 |    39 |     4  (25)| 00:00:01 |
|   1 |  SORT GROUP BY     |      |     3 |    39 |     4  (25)| 00:00:01 |
|   2 |   TABLE ACCESS FULL| VOTE |     6 |    78 |     3   (0)| 00:00:01 |
---------------------------------------------------------------------------

这是分析查询的计划：

Execution Plan
----------------------------------------------------------
Plan hash value: 781916126

---------------------------------------------------------------------------------
| Id  | Operation                | Name | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------------
|   0 | SELECT STATEMENT         |      |     6 |   288 |     4  (25)| 00:00:01 |
|*  1 |  VIEW                    |      |     6 |   288 |     4  (25)| 00:00:01 |
|*  2 |   WINDOW SORT PUSHED RANK|      |     6 |    78 |     4  (25)| 00:00:01 |
|   3 |    TABLE ACCESS FULL     | VOTE |     6 |    78 |     3   (0)| 00:00:01 |
---------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter("RN"=1)
   2 - filter(ROW_NUMBER() OVER ( PARTITION BY "ID_IDEA" ORDER BY
              INTERNAL_FUNCTION("DATE_VOTE") DESC  NULLS LAST)<=1)

此致罗布。

Answer 4

我通常使用first or last function执行此操作。它有一个奇怪的结构，可以解释为什么它不经常使用。请注意，只要order by子句是确定性的，那么max / min就不重要（但是需要因为这是构造函数的方式。

select 
  max(id) keep (dense_rank last order by date_vote) as id,
  id_idea,
  max(date_vote)
  from vote
group by id_idea

Oracle Group By Issue

4 个答案: