Sql 上述平均子查询未返回任何结果
我正试图得到一份期末考试成绩高于平均水平的学生名单 我首先选择平均值Sql 上述平均子查询未返回任何结果,sql,oracle,subquery,average,Sql,Oracle,Subquery,Average,我正试图得到一份期末考试成绩高于平均水平的学生名单 我首先选择平均值 SELECT w.LAST_NAME , AVG(s.NUMERIC_GRADE) AS NUMERIC_GRADE GRADE s , SECTION z, STUDENT w WHERE s.SECTION_ID = z.SECTION_ID AND s.STUDENT_ID = w.STUDENT_ID AND s.SECTION_ID = 90 AND s.GRADE_TYPE_CODE = 'FI' GROUP B
SELECT w.LAST_NAME , AVG(s.NUMERIC_GRADE) AS NUMERIC_GRADE
GRADE s , SECTION z, STUDENT w
WHERE s.SECTION_ID = z.SECTION_ID AND s.STUDENT_ID = w.STUDENT_ID
AND s.SECTION_ID = 90 AND s.GRADE_TYPE_CODE = 'FI'
GROUP BY w.LAST_NAME,s.NUMERIC_GRADE
我得到了这四个结果
LAST_NAME NUMERIC_GRADE
------------------------- -------------
Mulroy 83
Da Silva 92
Lopez 91
Abid 84
但是,当我试图从这四个查询中获得高于平均值的结果时,我没有得到任何行,看起来子查询和主查询具有相同的条件。我不知道如何做后平均以上
SELECT n.LAST_NAME , m.NUMERIC_GRADE
FROM GRADE m , STUDENT n
WHERE m.STUDENT_ID = n.STUDENT_ID
GROUP BY n.LAST_NAME , m.NUMERIC_GRADE
HAVING COUNT(*) >
(SELECT AVG (NUMERIC_GRADE)
FROM
(SELECT w.LAST_NAME , AVG(s.NUMERIC_GRADE) AS NUMERIC_GRADE
FROM GRADE s , SECTION z, STUDENT w
WHERE s.SECTION_ID = z.SECTION_ID AND s.STUDENT_ID = w.STUDENT_ID
AND s.SECTION_ID = 90 AND s.GRADE_TYPE_CODE = 'FI'
GROUP BY w.LAST_NAME,s.NUMERIC_GRADE))
ORDER BY n.LAST_NAME;
我想考91级和92级,因为它高于平均水平。当我试图选择期末考试成绩高于平均水平的学生时,为什么没有给我行?您的查询有几个问题:
- 您还需要在第一个
查询中使用groupby
AVG
- 您需要按照与内部查询相同的条件限制顶部查询(即等级类型代码和区段id)
- 您不需要加入部分,因为您正在限制其ID,该ID可从
表中获得GRADE
- 您应该使用至少与表名稍有相似的别名:这有助于提高可读性
- 为了更好的可读性,应该使用ANSI连接
SELECT n.LAST_NAME , AVG(m.NUMERIC_GRADE)
FROM GRADE g
JOIN STUDENT s ON g.STUDENT_ID = s.STUDENT_ID -- Use ANSI joins
WHERE g.SECTION_ID = 90 AND g.GRADE_TYPE_CODE = 'FI'
GROUP BY s.LAST_NAME
HAVING AVG(g.NUMERIC_GRADE) >
(SELECT AVG(NUMERIC_GRADE)
FROM (
SELECT AVG(g.NUMERIC_GRADE) AS NUMERIC_GRADE
FROM GRADE g
JOIN STUDENT s ON s.STUDENT_ID = g.STUDENT_ID
WHERE g.SECTION_ID = 90 AND g.GRADE_TYPE_CODE = 'FI'
GROUP BY s.LAST_NAME
)
)
ORDER BY s.LAST_NAME;