在Oracle Subquery中獲取訂單后的第一行

Question

我有一個表學生（身份證，姓名，部門，年齡，分數）。 我想找到每個系最高（最年輕的學生）得分最年輕的學生。 在SQL Server中，我可以使用以下SQL。

select * from student s1 
where s1.id in 
(select s2.id from student s2 
where s2.department = s1.department order by age asc, score desc top 1).

但是，在Oracle中，您不能在子查詢中使用order by子句，並且沒有limit / top like關鍵字。 我必須自己加入學生表兩次才能查詢結果。 在oracle中，我使用以下SQL。

select s1.* from student s1, 
(select s2.department, s2.age, max(s2.score) as max_score from student s2, 
(select s3.department, min(s3.age) as min_age from student s3 group by s3.department) tmp1 where 
s2.department = tmp1.department and s2.age = tmp1.min_age group by s2.department, s2.age) tmp2 
where s1.department =tmp2.department and s1.age = tmp2.age and s1.score=tmp2.max_score

有沒有人有任何想法為oracle簡化上述SQL。

Answer 1

試試這個

select * from
  (SELECT id, name, department, age, score,
  ROW_NUMBER() OVER (partition by department order by age desc, score asc) srlno 
  FROM student) 
where srlno = 1;

Answer 2

除了艾倫的回答，這也很好：

select * 
from (SELECT * 
  FROM student
  order by age asc, 
           score desc) 
where rownum = 1;

Answer 3

除了Bharat的回答之外，還可以在Oracle的子查詢中使用ORDER BY來做到這一點（正如Jeffrey Kemp指出的那樣）：

SELECT *
FROM   student s1
WHERE  s1.id IN (SELECT id
                 FROM   (SELECT   id, ROWNUM AS rn
                         FROM     student s2
                         WHERE    s1.department = s2.department
                         ORDER BY age ASC, score DESC)
                 WHERE  rn = 1);

如果您使用此方法，您可能想要刪除子查詢，只需使用rownum = 1 。 這將導致不正確的結果，因為排序將在標准之后應用（您將獲得已排序的1行，而不是排序集中的一行）。

Answer 4

select to_char(job_trigger_time,'mm-dd-yyyy') ,job_status from
(select * from kdyer.job_instances ji INNER JOIN kdyer.job_param_values pm 
on((ji.job_id = pm.job_id) and (ji.job_spec_id = '10003') and (pm.param_value='21692') )
order by ji.job_trigger_time desc)
where rownum<'2'