SQL - 存在比較

Question

我需要選擇一個國家有一個處理行的行，並且同一個國家存在一個未處理的行。對於下表，我將返回與id = 1行。

所以首先我需要檢查是否有一排TREATED狀態，然後我必須檢查是否有一個UNTREATED狀態的行與TREATED狀態相同的國家。

Table example 1 
    +----+--------+--------------+ 
    | id | country| status  | 
    +----+--------+--------------+ 
    | 1 | SE  | TREATED  | 
    | 2 | DK  | UNTREATED | 
    | 3 | SE  | UNTREATED | 
    +----+--------+--------------+ 

    Result of query 
    +----+--------+--------------+ 
    | id | country| status  | 
    +----+--------+--------------+ 
    | 1 | SE  | TREATED  | 
    +----+--------+--------------+ 

如果不存在未經處理的行或該國不一樣，那麼我們不應該返回任何

Table example 2 
    +----+--------+--------------+ 
    | id | country| status  | 
    +----+--------+--------------+ 
    | 1 | SE  | TREATED  | 
    | 2 | DK  | UNTREATED | 
    | 3 | US  | UNTREATED | 
    +----+--------+--------------+ 

    Result of query (empty) 
    +----+--------+--------------+ 
    | id | country| status  | 
    +----+--------+--------------+ 

Answer 1

這裏有一對夫婦的其他選項：

select t1.id,t1.country,t1.status 
    from table t1 
    join table t2 on t1.country=t2.country and t2.status='UNTREATED' 
    where t1.status='TREATED' 


select id,country,status 
      from table t1 
      where 
       t1.status='TREATED' and 
       exists (
         select 1 from table t2 where 
          t2.country = t1.country and 
          t2.status = 'UNTREATED' 
       ) 

我相信第一個，使用一個連接，可能會表現最佳。

Answer 2

Select country 
from table 
where status in ('treated', 'untreated') -- only needed if there are other statuses 
group by country 
having count(distinct status) = 2 

Answer 3

SELECT country FROM table 
WHERE status IN ('treated', 'untreated') 
GROUP BY country HAVING count(DISTINCT status)=2; 

語法可能會因SQL有所不同，不知道約db2。