Question

我有如下数据。我只想选择一个优先级较高的task_id。

TASK_ID : 13912 = 2 rows
          13913 = 3 rows

Task_ID owner   priority
13912   JUNGIL  99
13912   BANCS   1
13913   JUNGIL  99
13913   BANCS   1
13913   AIL     2

还有更多数据，但我必须只选择一个。

Answer 1

keep ... dense_rank first（thanks for the reminder Gordon Linoff）的一些急需的练习：

select task_id,
  max(owner) keep (dense_rank last order by priority) as owner,
  max(priority) keep (dense_rank last order by priority) as priority
from t42
group by task_id
order by task_id;

   TASK_ID OWNER        PRIORITY
---------- ---------- ----------
     13912 JUNGIL             99 
     13913 JUNGIL             99

如果最高优先级为1则为：

select task_id,
  max(owner) keep (dense_rank first order by priority) as owner,
  max(priority) keep (dense_rank first order by priority) as priority
from t42
group by task_id
order by task_id;

   TASK_ID OWNER        PRIORITY
---------- ---------- ----------
     13912 BANCS               1 
     13913 BANCS               1

您也可以使用子查询和分析来执行此操作：

select task_id, owner, priority
from (
  select task_id, owner, priority,
    dense_rank() over (partition by task_id order by priority) as rnk
  from t42
)
where rnk = 1
order by task_id;

   TASK_ID OWNER        PRIORITY
---------- ---------- ----------
     13912 BANCS               1 
     13913 BANCS               1

...但keep方法更简单。 SQL Fiddle显示两者。

Answer 2

您似乎只需要group by这样的查询：

select max(priority), task_id
  from <YOUR_TABLE_HERE>
 group by t.task_id

如果你想拥有所有行的列

select t.*
  from t
  join (
    select max(priority) priority, task_id
      from t
     group by t.task_id
  )t2 on t.task_id = t2.task_id and t.priority = t2.priority

删除选择查询中的重复行

2 个答案: