查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004） [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

SELECT DISTINCT top 3 SC.S# As 学生学号, Student.Sname AS 学生姓名 , T1.score AS 企业管理, T2.score AS 马克思, T3.score AS UML, T4.score AS 数据库, ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分 FROM Student,SC LEFT JOIN SC AS T1 ON SC.S# = T1.S# AND T1.C# = ‘001’ LEFT JOIN SC AS T2 ON SC.S# = T2.S# AND T2.C# = ‘002’ LEFT JOIN SC AS T3 ON SC.S# = T3.S# AND T3.C# = ‘003’ LEFT JOIN SC AS T4 ON SC.S# = T4.S# AND T4.C# = ‘004’ WHERE student.S#=SC.S# and ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) NOT IN (SELECT DISTINCT TOP 15 WITH TIES ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) FROM sc LEFT JOIN sc AS T1 ON sc.S# = T1.S# AND T1.C# = ‘k1’ LEFT JOIN sc AS T2 ON sc.S# = T2.S# AND T2.C# = ‘k2’ LEFT JOIN sc AS T3 ON sc.S# = T3.S# AND T3.C# = ‘k3’ LEFT JOIN sc AS T4 ON sc.S# = T4.S# AND T4.C# = ‘k4’ ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);