Oracle Dense_rank

Question

我需要根据ID和时间删除重复的行column.I需要保留最新时间的记录only.If有两条记录的最大时间，我可以保留任何一条记录并删除其中的所有其他记录 group.Please 在下面找到我的输入数据

ID  TIMES
123 13/01/2018
123 14/01/2018
123 15/01/2018
345 14/01/2018
567 20/01/2018
567 20/01/2018
879 NULL
879 21/01/2018

我已经为 same.But 编写了一个查询，它不适用于 ID=567 的情况，因为它们在时间列中具有相同的值。请在下面找到我的查询

delete FROM table where (ID,times) in( 
  SELECT ID,times, 
    RANK() OVER (PARTITION BY ID ORDER BY times DESC NULLS LAST) dest_rank
    FROM TABLE 
  ) WHERE dest_rank <> 1

有什么办法可以做到这一点。

Answer 1

这是一种方法：

delete t from t
    where rowid <> (select max(rowid) keep (dense_rank first order by times desc)
                    from t t2
                    where t2.id = t.id
                   );

不过，我会用一个临时的 table:

create temporary table tt
    select id, max(times) as times
    from t
    group by id;

truncate table t;

insert into t(id, times)
    select id, times
    from tt;

Answer 2

您可以通过贡献或rowid :

获得成功

delete mytable where (rowid) in
(
  select t1.rowid from mytable t1
   where times <
  (
  select max(times)
    from mytable t2
   where t2.id = t1.id  
     and t2.times != t1.times -- for non-matching records of times
  )
  union all
  select t1.rowid from mytable t1
   where rowid <
  (
  select max(rowid)
    from mytable t2
   where t2.id = t1.id
     and t2.times = t1.times  -- for matching records of times
  )
);

Answer 3

我会

delete demo where rowid in
( select lag(rowid) over (partition by id order by times nulls first) from demo  );

您没有说明您希望如何处理空值。如果您想保留空日期的行，请将 nulls first 更改为 nulls last。

Oracle Dense_rank

Oracle Dense Rank

sql

oracle

dml