sql - 子查询 - 获得最高分

Question

我正在努力寻找期末考试得分最高的学生

首先我选择

SELECT s.STUDENT_ID, w.LAST_NAME,w.FIRST_NAME, MAX(s.NUMERIC_GRADE) AS NUMERIC_FINAL_GRADE
FROM GRADE s , SECTION z, STUDENT w
WHERE s.SECTION_ID = z.SECTION_ID AND s.STUDENT_ID = w.STUDENT_ID
AND z.COURSE_NO = 230 AND z.SECTION_ID = 100 AND s.GRADE_TYPE_CODE = 'FI'
GROUP BY s.STUDENT_ID, w.FIRST_NAME,w.LAST_NAME

它给了我这个结果，这就是我想要的

 STUDENT_ID LAST_NAME                 FIRST_NAME                NUMERIC_FINAL_GRADE
  ---------- ------------------------- ------------------------- -------------------
   262 Walston                   Donna                                      85 
   141 Boyd                      Robert                                     84 

但是当我尝试从这两个中获取最大值时，它没有给我任何行或错误

i.STUDENT_ID, k.LAST_NAME,k.FIRST_NAME
FROM GRADE i , SECTION j, STUDENT k
WHERE i.SECTION_ID = j.SECTION_ID AND i.STUDENT_ID = k.STUDENT_ID
AND j.COURSE_NO = 230 AND j.SECTION_ID = 100 AND i.GRADE_TYPE_CODE = 'FI' 
GROUP BY i.STUDENT_ID, k.FIRST_NAME,k.LAST_NAME
HAVING COUNT(*) =
(SELECT MAX(NUMERIC_FINAL_GRADE)
 FROM
(SELECT s.STUDENT_ID, w.LAST_NAME,w.FIRST_NAME, MAX(s.NUMERIC_GRADE) AS NUMERIC_FINAL_GRADE
FROM GRADE s , SECTION z, STUDENT w
WHERE s.SECTION_ID = z.SECTION_ID AND s.STUDENT_ID = w.STUDENT_ID
AND z.COURSE_NO = 230 AND z.SECTION_ID = 100 AND s.GRADE_TYPE_CODE = 'FI'
GROUP BY s.STUDENT_ID, w.FIRST_NAME,w.LAST_NAME))

ORDER BY i.STUDENT_ID, k.LAST_NAME,k.FIRST_NAME;

如何从我已有的这两个结果中获得最大结果以及为什么它没有给我任何行或错误？

Answer 1

传统的方法是analytic MAX() (或其他分析函数):

select *
  from ( select s.student_id
              , w.last_name
              , w.first_name
              , s.numeric_grade
              , max(s.numeric_grade) over () as numeric_final_grade
           from grade s
           join section z
             on s.section_id = z.section_id
           join student w
             on s.student_id = w.student_id
          where z.course_no = 230 
            and z.section_id = 100 
            and s.grade_type_code = 'FI'
                )
 where numeric_grade = numeric_final_grade

但我可能更喜欢使用 FIRST (保持)。

select max(s.student_id) keep (dense_rank first order by s.numeric_grade desc) as student_id
     , max(w.last_name) keep (dense_rank first order by s.numeric_grade desc) as last_name
     , max(w.first_name) keep (dense_rank first order by s.numeric_grade desc) as first_na,e
     , max(s.numeric_grade_name) as numeric_final_grade
  from grade s
  join section z
    on s.section_id = z.section_id
  join student w
    on s.student_id = w.student_id
 where z.course_no = 230 
   and z.section_id = 100 
   and s.grade_type_code = 'FI'

与您最初建议的方法相比，这两种方法的好处是您只需扫描表一次，无需再次访问表或索引。我强烈推荐Rob van Wijk's blog post论两者的区别。

附注这些将返回不同的结果，因此它们略有不同。如果两个学生具有相同的最高分数，分析函数将保留重复项(这也是您的建议所做的)。聚合函数将删除重复项，如果出现平局则返回随机记录。