Question

我需要根据ID和时间列删除重复的行。我需要保留最新时间的记录。如果有两个记录有最大次数，我可以保留任何记录并删除其中的所有其他记录那个组。请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

我已经为同一个写了一个查询。但它没有使用ID = 567的情况，因为它们在时间列中都有相同的值。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

我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
                   );

但是，我会用临时表来执行此操作：

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

你可以通过控制或rowid：

取得成功

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  );

您没有说明如何处理null值。如果要保留具有空日期的行，请将nulls first更改为nulls last。

Oracle密集排名

3 个答案: