SQL 经典题型解答（5）

@(数据库)

20、查询学生的总成绩并进行排名

SELECT
    a.s,
    a.Sname,
    SUM( b.score ) AS sumscore,
    RANK() OVER (ORDER BY SUM( b.score ) DESC) AS ‘名次‘
FROM
    student a,
    sc b
WHERE
    a.s = b.S
GROUP BY
    a.S,a.Sname

详解：

没有用到新的知识。

程序运行结果：

SELECT
    a.C,
    a.Cname,
    c.Tname,
    avg( b.score ) AS avgscore
FROM
    course a,
    sc b,
    teacher C
WHERE
    a.C = b.C
    AND a.T = c.T
GROUP BY
    a.c,
    a.Cname,
    c.Tname
ORDER BY
    avgscore DESC

程序运行结果：

SELECT
    a.*,
    b.c,
    b.score
FROM
    student a,
    (
    SELECT
        *
    FROM
        ( SELECT s, c, score, ROW_NUMBER ( ) OVER ( ORDER BY score DESC ) AS mc FROM sc WHERE C = ‘01‘ ) a
    WHERE
        mc BETWEEN 2
        AND 3 UNION ALL
    SELECT
        *
    FROM
        ( SELECT s, c, score, ROW_NUMBER ( ) OVER ( ORDER BY score DESC ) AS mc FROM sc WHERE C = ‘02‘ ) b
    WHERE
        mc BETWEEN 2
        AND 3 UNION ALL
    SELECT
        *
    FROM
        ( SELECT s, c, score, ROW_NUMBER ( ) OVER ( ORDER BY score DESC ) AS mc FROM sc WHERE C = ‘03‘ ) c
    WHERE
        mc BETWEEN 2
        AND 3
    ) b
WHERE
    a.s = b.s
ORDER BY
    c

详解；

通过查询表 sc 可以根据每一个课程的成绩生成表格。然后查出排名在 2 和 3 之间的学生的学号、课程、以及课程成绩，在联合表 stuedent 得到血色和功能的信息。

由于课程较少，可以采用这种方法，但是如果课程数多就不可以，应该采用 SQL 循环，

如果题目要求中没有声明选取课程几，代码中就不应该出现 ’01‘，’02‘，’03’ 这样的课程。

SQL UNION 操作符用法

程序运行结果：

原文地址：https://www.cnblogs.com/wobu/p/9637601.html