SQL 经典题型解答

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SQL 经典习题解答(6)

23、统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比

SELECT
	t1.*,
	t2.all_num,
	CONCAT( ROUND( t1.num / t2.all_num * 100, 2 ), \'%\' ) \'百分比\'
FROM
	(
	SELECT
		m.C,
		m.Cname,
		(
		CASE
				
				WHEN n.score >= 85 THEN
				\'85-100\' 
				WHEN n.score >= 70 
				AND n.score < 85 THEN \'70-85\' WHEN n.score >= 60 
					AND n.score < 70 THEN
						\'60-70\' ELSE \'0-60\' 
					END 
					) AS px,
					count( 1 ) num 
				FROM
					Course m,
					sc n 
				WHERE
					m.C = n.C 
				GROUP BY
					m.C,
					m.Cname,
					px 
				ORDER BY
					m.C 
				) t1,
				(
				SELECT
					m.C,
					m.Cname,
					count( 1 ) all_num 
				FROM
					Course m,
					sc n 
				WHERE
					m.C = n.C 
				GROUP BY
					m.C,
					m.Cname 
				ORDER BY
					m.C 
				) t2 
		WHERE
	t1.c = t2.c

详解:

首先统计各科成绩各分数段人数:课程编号,课程名称,选择表 sc 和表 course ,通过 CASE ... WHEN ... THEN ... ELSE ... END 语句分出分数段,再查出每一个课程学习的总人数,最后相除即可得到百分比。
CASE ... WHEN ... THEN ... ELSE ... END 用法参考
SQL 字符串拼接

程序运行结果:


24、查询学生平均成绩及其名次

SELECT
	a.*,
	b.avgscore,
	b.mc 
FROM
	student a,
	(
	SELECT
		s,
		avg( score ) AS avgscore,
		rank ( ) over ( ORDER BY avg( score ) DESC ) AS mc 
	FROM
		sc 
	GROUP BY
		S 
	) b 
WHERE
	a.s = b.s 
ORDER BY
	mc

详解:

首先从表 sc 中查出每个学生的平均成绩和根据平均成绩进行的排名,再与表 student 连接得到结果

程序运行结果:

25、查询各科成绩前三名的记录

SELECT
	a.*,
	b.c,
	b.score,
	b.mc 
FROM
	student a,
	( SELECT *, row_number ( ) over ( PARTITION BY c ORDER BY score DESC ) AS mc FROM sc ) b 
WHERE
	a.s = b.s 
	AND mc BETWEEN 1 
	AND 3 
ORDER BY
	c,
	mc

详解:

首先在表 sc 根据课程成绩生成每一门课程的排名记为表 b ,然后与表 student 连接得到结果

程序运行结果:


26、查询每门课程被选修的学生数

SELECT
	c,
	count( s ) AS num 
FROM
	sc 
GROUP BY
	c

程序运行结果:


27、查询出只有两门课程的全部学生的学号和姓名

SELECT 
	a.s,
	a.sname 
FROM
	student a,
	( SELECT s FROM sc GROUP BY s HAVING count( s ) = 2 ) b 
WHERE
	a.s = b.s

详解:

在表 sc 中,学号出现的次数即为学生课程数,通过 GROUP BYHAVING 函数得出选课数为 2 的学生学号,连接表 student 得出结果

程序运行结果:


28、查询男生、女生人数

SELECT	Ssex,count(s) FROM student WHERE Ssex = \'男\'
UNION ALL
SELECT	Ssex,count(s) FROM student WHERE Ssex = \'女\'

程序运行结果:


29、查询名字中含有"风"字的学生信息

SELECT
	* 
FROM
	student 
WHERE
	Sname LIKE \'%风%\'

程序运行结果:


30、查询同名同性学生名单,并统计同名人数

SELECT
	Sname,
	Ssex,
	COUNT( 1 ) num 
FROM
	student 
GROUP BY
	Sname,
	Ssex 
HAVING
	count( 1 ) > 1

详解:

通过 GROUP BY 划分出同名同性的学生,在通过 HAVING 判断人数是否大于 1
程序运行结果:


31、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime)

SELECT
	* 
FROM
	student 
WHERE
	Sage LIKE \'1990%\'

程序运行结果:


32、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号

SELECT
	c,
	avg( score ) AS avgscore 
FROM
	sc 
GROUP BY
	c 
ORDER BY
	avg( score ) DESC,
	c 

详解:

ORDER BY,先根据 avg( score )排序,如果平均成绩相同,再根据课程编号升序排列

程序运行结果:


33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

SELECT
	a.s,
	a.sname,
	b.avgscore 
FROM
	student a,
	( SELECT s, avg( score ) AS avgscore FROM sc GROUP BY s HAVING avg( score ) >= 85 ) b
WHERE a.s = b.s

程序运行结果:


34、查询课程名称为"数学",且分数低于60的学生姓名和分数

SELECT
	a.sname,
	b.score 
FROM
	student a,
	sc b,
	course c 
WHERE
	a.s = b.s 
	AND b.c = c.C 
	AND b.score < 60 
	AND c.Cname = \'数学\'

程序运行结果:

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