[英]Above AVG sub-query returning no results
我正在尝试列出期末考试成绩高于平均水平的学生
我首先选择平均
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
我得到这四个结果
LAST_NAME NUMERIC_GRADE
------------------------- -------------
Mulroy 83
Da Silva 92
Lopez 91
Abid 84
但是,当我尝试从这四个中获得上述平均值时,我没有任何行,并且看起来子查询和主查询具有相同的条件。 我不确定在平均值之后如何执行上述操作。
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;
我想获得numberic_grade 91和92,因为它高于平均值。 为什么当我试图选择那些期末考试中的成绩高于平均水平的人时,为什么没有任何行?
您的查询存在几个问题:
GROUP BY
查询中使用AVG
GRADE
表中获得该ID。 尝试查询以下更正:
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;
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