Question

假设您有一个表格，如：

id   foreign_key  status
------------------------
1    1            new
2    1            incative
3    1            approved
4    2            new
5    2            new
6    2            approved
7    3            new
8    3            approved
9    4            approved

如何找到记录，其中给定的foreign_key只有一个记录处于状态new而另一个记录被批准，就像foreign_key 3一样？

Answer 1

select foreign_key from table
group by foreign_key
having 
   abs(1 - count(case status when 'new' then 1 end)) + 
   abs(count(1) - 1 - count(case status when 'approved' then 1 end)) = 0

Answer 2

像这样的东西

Select *
from 
(
   Select foreign_key
     from table
    where status = 'new'
    group by foreign_key
    having count(1) = 1
) new_st
inner join
(
   Select foreign_key
     from table
    where status = 'approved'
    group by foreign_key
    having count(1) = (select count(1)-1 from table t1 where t1.foreign_key =foreign_key)
) app_st
on new_st.foreign_key = app_st.foreign_key

Answer 3

SELECT *
  FROM (SELECT id, foreign_key, status,
               COUNT (DECODE (status, 'new', 1))
                  OVER (PARTITION BY foreign_key)
                  new_count,
               COUNT (DECODE (status, 'approved', 1))
                  OVER (PARTITION BY foreign_key)
                  approved_count,
               COUNT (status) OVER (PARTITION BY foreign_key) total_count
          FROM mytable)
 WHERE new_count = 1 AND new_count + approved_count = total_count;

我使用了3种不同的计数。一个计数新的，一个计数批准，一个计数所有状态。最后，只选择new_count = 1且new_count + approved_count等于total_count的记录。

演示here。

编辑：可以添加approved_count > 0条件，以确保至少有一个已批准的状态。

如何在Oracle中找到相关值？

3 个答案: