SQL必会50题

根据知乎进行练习, 有些还是有点难度

链接https://zhuanlan.zhihu.com/p/43289968

-- 1.查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号（重点）
USE sql_test;
SELECT a.s_id, a.s_score AS '01', b.s_score AS '02', s.s_name
FROM 
(SELECT s_id, c_id, s_score FROM score WHERE c_id = 01) AS a
JOIN
(SELECT s_id, c_id, s_score FROM score WHERE c_id = 02) AS b
ON a.s_id = b.s_id
JOIN student s
ON a.s_id = s.s_id
WHERE a.s_score > b.s_score;
-- 2、查询平均成绩大于60分的学生的学号和平均成绩（简单，第二道重点）
SELECT sc.s_id, st.s_name, AVG(sc.s_score) AS avg_score
FROM score sc
JOIN student st
ON sc.s_id = st.s_id
GROUP BY s_id
HAVING avg_score > 60;

-- 3、查询所有学生的学号、姓名、选课数、总成绩（不重要）
SELECT 
	st.s_id,
    st.s_name,
    COUNT(sc.c_id),
    SUM(IF(sc.s_score IS NULL, 0, sc.s_score))
FROM student st
LEFT JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id;

SELECT 
	st.s_id,
    st.s_name,
    COUNT(sc.c_id),
    SUM(CASE WHEN sc.s_score IS NULL THEN 0 ELSE sc.s_score END)
FROM student st
LEFT JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id;

-- 4、查询姓“猴”的老师的个数（不重要）
SELECT COUNT(*)
FROM teacher
WHERE t_name LIKE '猴%';

-- 5、查询没学过“张三”老师课的学生的学号、姓名（重点）
SELECT *
FROM student st
WHERE st.s_id NOT IN
	(SELECT sc.s_id
	FROM course c
		JOIN score sc USING(c_id)
	WHERE c.t_id =
		(SELECT t_id
		FROM teacher
		WHERE t_name = '张三'));
-- 6、查询学过“张三”老师所教的所有课的同学的学号、姓名（重点）
USE sql_test;
SELECT *
FROM student st
WHERE st.s_id  IN
	(SELECT sc.s_id
	FROM course c
		JOIN score sc USING(c_id)
	WHERE c.t_id =
		(SELECT t_id
		FROM teacher
		WHERE t_name = '张三'));
        
-- 7、查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名（重点）
SELECT 
	a.s_id,
    st.s_name
FROM 
	(SELECT *
	FROM score 
	WHERE c_id = '01') AS a
JOIN 
	(SELECT *
	FROM score
	WHERE c_id = '02') AS b
On a.s_id = b.s_id
JOIN student st
ON a.s_id = st.s_id;

SELECT *
FROM student
WHERE s_id IN (
	(SELECT a.s_id FROM 
	(SELECT * FROM score WHERE c_id = '01') AS a
	JOIN 
	(SELECT * FROM score WHERE c_id = '02') AS b
	On a.s_id = b.s_id)
);

-- 8、查询课程编号为“02”的总成绩（不重点）
SELECT SUM(s_score)
FROM score
WHERE c_id = '02';
-- 9、查询所有课程成绩小于60分的学生的学号、姓名
-- 也可以用子查询
SELECT 
	DISTINCT st.s_id, 
    st.s_name
FROM score sc
JOIN student st
ON sc.s_id = st.s_id
WHERE s_score < 60;
-- 10.查询没有学全所有课的学生的学号、姓名(重点)
-- 这个答案不全, 主要是从成绩表查询,但是有些没选课,因此没有成绩
SELECT * 
FROM student st
JOIN 
	(SELECT s_id FROM score
	GROUP BY s_id 
	HAVING count(DISTINCT c_id) < (SELECT COUNT( DISTINCT c_id) FROM course)) AS a
ON st.s_id = a.s_id;
-- 反过来思考, 所有学完课程的学生比较好查询
SELECT 
	st.s_id, 
    st.s_name,
    COUNT(DISTINCT sc.c_id) AS count_c
FROM student st
LEFT JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id
HAVING count_c < 3;
-- 11、查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名（重点）
SELECT *
FROM student st
JOIN (
	SELECT DISTINCT s_id
	FROM score 
	WHERE c_id IN (
		SELECT c_id 
		FROM score
		WHERE s_id = '01'
	) AND s_id != '01') AS sc 
ON st.s_id = sc.s_id;

-- 12.查询和“01”号同学所学课程完全相同的其他同学的学号(重点)

SELECT * FROM student
WHERE s_id IN (
	SELECT s_id 
	FROM score
	WHERE s_id != '01'
	GROUP BY s_id 
	HAVING COUNT(DISTINCT c_id) = (
		SELECT COUNT(DISTINCT c_id)
		FROM score
		WHERE s_id = '01')
)
AND s_id NOT IN(
	SELECT s_id FROM score
	WHERE c_id NOT IN (
		SELECT c_id
		FROM score
		WHERE s_id = '01'
	)
);
--  13、查询没学过"张三"老师讲授的任一门课程的学生姓名 和47题一样（重点，能做出来）
SELECT * FROM student st
WHERE st.s_id NOT IN (
	SELECT sc.s_id
	FROM score sc
	WHERE sc.c_id = (
		SELECT c_id FROM course c
		WHERE c.t_id = (
			SELECT t_id
			FROM teacher
			WHERE t_name = '张三'))
);
-- 15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩（重点）

SELECT sc.s_id,
	AVG(sc.s_score),
    st.s_name
FROM score sc
JOIN student st
ON sc.s_id = st.s_id
WHERE sc.s_score < 60 
GROUP BY sc.s_id
HAVING COUNT(sc.c_id) >= 2;

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息（和34题重复，不重点）

SELECT *
FROM student st
JOIN score sc
ON st.s_id = sc.s_id
WHERE sc.s_score < 60 AND sc.c_id = '01'
ORDER BY sc.s_score DESC;
-- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
-- 格式不清晰

SELECT s_id,
	st.s_id,
    sc.c_id,
    sc.s_score,
    avg_score
FROM student st
LEFT JOIN score sc
ON st.s_id = sc.s_id
LEFT JOIN 
	(SELECT 
	sc.s_id,
    AVG(sc.s_score) as avg_score
	FROM score sc
	GROUP BY sc.s_id) AS avg_s
ON st.s_id = avg_s.s_id
ORDER BY avg_score DESC;
-- 第二种结果
USE sql_test;
SELECT 
	s_id,
    MAX(CASE WHEN c_id = '01' THEN s_score ELSE NULL END) '语文',
    MAX(CASE WHEN c_id = '02' THEN s_score ELSE NULL END) '数学',
    MAX(CASE WHEN c_id = '03' THEN s_score ELSE NULL END) '英语',
    AVG(s_score) 
FROM score sc
GROUP BY s_id
ORDER BY AVG(s_score) DESC;

-- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

-- 及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90 (超级重点)

SELECT 
	sc.c_id,
    c.c_name,
    MAX(sc.s_score),
    MIN(sc.s_score),
    AVG(sc.s_score),
    SUM(CASE WHEN sc.s_score >= 60 THEN 1.0 ELSE 0 END) / COUNT(s_id) '及格率',
    AVG(CASE WHEN sc.s_score >= 70 AND sc.s_score <80 THEN 1.0 ELSE 0 END) '中等率',
    AVG(CASE WHEN sc.s_score >= 80 AND sc.s_score <90 THEN 1.0 ELSE 0 END) '优良率',
    AVG(CASE WHEN sc.s_score >= 90 THEN 1.0 ELSE 0 END) '优秀率'
FROM score sc
JOIN course c
ON sc.c_id = c.c_id
GROUP BY c_id;

-- 19、按各科成绩进行排序，并显示排名(重点row_number)

SELECT 
	s_id,
    c_id,
    RANK() OVER(PARTITION BY c_id ORDER BY s_score)
FROM score;

-- 查询学生的总成绩并进行排名（不重点）, 一般都是用RANK
-- ROW_NUMBER
-- RANK
-- DENSE_RANK
USE sql_test;
SELECT 
	s_id,
    SUM(s_score) AS sum_score
FROM score
GROUP BY s_id
ORDER BY sum_score DESC;
-- 21 、查询不同老师所教不同课程平均分从高到低显示(不重点)
-- 以课程为主体
SELECT 
	c.c_id,
    c.c_name,
    AVG(s_score)
FROM course c
JOIN score sc
ON c.c_id = sc.c_id
GROUP BY c.c_id
ORDER BY  AVG(s_score);
-- 以老师为主体
SELECT 
	te.t_id,
    te.t_name,
    AVG(s_score)
FROM course c
JOIN score sc
ON c.c_id = sc.c_id
JOIN teacher te
ON te.t_id = c.t_id
GROUP BY c.c_id
ORDER BY  AVG(s_score) DESC;

-- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩（重要 25类似）

SELECT *
FROM (
	SELECT 
		st.s_id,
		st.s_name,
		st.s_birth,
		st.s_sex,
		c_id,
		s_score,
		ROW_NUMBER () OVER (PARTITION BY c_id ORDER BY s_score DESC) m
	FROM score sc
	JOIN student st
	ON sc.s_id = st.s_id) a
WHERE m IN (2, 3);

-- 23、使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数：课程ID和课程名称(重点和18题类似)
SELECT 
	sc.c_id,
    c.c_name,
    SUM(CASE WHEN s_score < 60 THEN 1 ELSE 0 END) '<60',
    COUNT(CASE WHEN s_score <= 70 AND s_score > 60 THEN 1 ELSE NULL END) '70-60',
    SUM(CASE WHEN s_score <= 85 AND s_score > 70 THEN 1 ELSE 0 END) '85-70',
	SUM(CASE WHEN s_score > 85 THEN 1 ELSE 0 END) '100-85'
FROM score sc
JOIN course c
ON sc.c_id = c.c_id
GROUP BY c_id;

-- 24、查询学生平均成绩及其名次（同19题，重点）

SELECT 
	s_id,
    AVG(s_score),
    ROW_NUMBER() OVER(PARTITION BY s_id ORDER BY AVG(s_score)) m
FROM score
GROUP BY s_id;

-- 25、查询各科成绩前三名的记录（不考虑成绩并列情况）（重点 与22题类似）
SELECT *
FROM (
	SELECT 
		st.s_id,
		st.s_name,
		st.s_birth,
		st.s_sex,
		c_id,
		s_score,
		ROW_NUMBER () OVER (PARTITION BY c_id ORDER BY s_score DESC) m
	FROM score sc
	JOIN student st
	ON sc.s_id = st.s_id) a
WHERE m IN (1, 2, 3);
-- 26、查询每门课程被选修的学生数(不重点)
SELECT 
	c.c_id,
    c.c_name,
    COUNT(sc.s_id)
FROM course c
JOIN score sc
ON c.c_id = sc.c_id
GROUP BY c.c_id;

-- 27、 查询出只有两门课程的全部学生的学号和姓名(不重点)
SELECT 
	st.s_id,
    st.s_name,
    COUNT(sc.c_id) as count_c
FROM student st
JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id
HAVING count_c = 2;

-- 28、查询男生、女生人数(不重点)
SELECT 
	s_sex,
    COUNT(s_id)
FROM student
GROUP BY s_sex;

-- 29 查询名字中含有"风"字的学生信息（不重点）
SELECT *
FROM student
WHERE s_name LIKE '%风%';
-- 31、查询1990年出生的学生名单（重点year）

SELECT *
FROM student 
WHERE YEAR(s_birth) = 1990;

-- 32、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
USE sql_test;
SELECT 
	st.s_id,
    st.s_name,
    AVG(sc.s_score)
FROM student st
JOIN score sc
ON st.s_id = sc.s_id
GROUP BY st.s_id
HAVING AVG(sc.s_score) > 85;

-- 33、查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列（不重要）
SELECT 
	c_id,
    AVG(s_score)
FROM score
GROUP BY c_id
ORDER BY AVG(s_score), c_id DESC;

-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数（不重点）
SELECT 
	st.s_name,
    sc.s_score
FROM course co
JOIN score sc ON co.c_id = sc.c_id
JOIN student st ON sc.s_id = st.s_id
WHERE c_name = '数学' AND sc.s_score < 60;

-- 35查询所有学生的课程及分数情况（重点）
SELECT 
	sc.s_id,
    MAX(CASE WHEN co.c_name = '数学' THEN sc.s_score ELSE NULL END) '数学',
    MAX(CASE WHEN co.c_name = '语文' THEN sc.s_score ELSE NULL END) '语文',
    MAX(CASE WHEN co.c_name = '英语' THEN sc.s_score ELSE NULL END) '英语'
FROM score sc
JOIN course co ON sc.c_id = co.c_id
GROUP BY sc.s_id;

-- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数（重点）

SELECT 
	sc.s_id,
	st.s_name,
    sc.s_score
FROM score sc
JOIN student st ON sc.s_id = st.s_id
WHERE sc.s_score > 70;
-- 37、查询不及格的课程并按课程号从大到小排列(不重点)
SELECT *
FROM score
WHERE s_score < 60
ORDER BY c_id DESC;
-- 38、查询课程编号为03且课程成绩在80分以上的学生的学号和姓名
SELECT 
	sc.s_id,
	st.s_name
FROM score sc
JOIN student st ON sc.s_id = st.s_id
WHERE sc.c_id = '03' AND sc.s_score > 80;
-- 39、求每门课程的学生人数（不重要）
SELECT 
	c_id,
	COUNT(c_id)
FROM score
GROUP BY c_id;


-- 40、查询选修“张三”老师所授课程的学生中成绩最高的学生姓名及其成绩（重要top）
SELECT 
	st.s_name,
    sc.s_score
FROM teacher te
JOIN course co ON te.t_id = co.t_id
JOIN score sc ON co.c_id = sc.c_id
JOIN student st ON sc.s_id = st.s_id
WHERE te.t_name = '张三'
ORDER BY sc.s_score DESC LIMIT 1;

-- 41.查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 （重点）
USE sql_test;
SELECT 
	DISTINCT s1.s_id, s1.c_id, s1.s_score
FROM score s1
JOIN score s2
ON s1.s_id = s2.s_id
WHERE s1.s_score = s2.s_score AND s1.c_id != s2.c_id;

-- 42查询每门功成绩最好的前两名
SELECT *
FROM (
	SELECT 
		st.s_id,
		st.s_name,
		st.s_birth,
		st.s_sex,
		c_id,
		s_score,
		ROW_NUMBER () OVER (PARTITION BY c_id ORDER BY s_score DESC) m
	FROM score sc
	JOIN student st
	ON sc.s_id = st.s_id) a
WHERE m IN (1, 2);

-- 43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列（不重要）
SELECT 
	c_id,
    COUNT(c_id)
FROM score
GROUP BY c_id
HAVING COUNT(c_id) > 5
ORDER BY COUNT(c_id) DESC, c_id;

-- 44、检索至少选修两门课程的学生学号（不重要）
SELECT 
	s_id,
    COUNT(DISTINCT c_id) AS count_course
FROM score
GROUP BY s_id
HAVING count_course >= 2;

-- 45、 查询选修了全部课程的学生信息（重点划红线地方）
SELECT 
	s_id,
    COUNT(c_id)
FROM score
GROUP BY s_id
HAVING COUNT(c_id) = (SELECT COUNT(*) FROM course);

-- 47、查询没学过“张三”老师讲授的任一门课程的学生姓名（还可以，自己写的，答案中没有）
SELECT *
FROM student
WHERE s_id NOT IN (
	SELECT sc.s_id
	FROM score sc
	JOIN course co ON sc.c_id = co.c_id
	JOIN teacher te ON co.t_id = te.t_id
	WHERE te.t_name = '张三'
);
-- 查询两门以上不及格课程的同学的学号及其平均成绩（还可以，自己写的，答案中没有）

SELECT 
	s_id, 
	COUNT(s_score), 
    AVG(s_score)
FROM score
WHERE s_score < 60
GROUP BY s_id
HAVING COUNT(s_score) >= 2;

-- 46、查询各学生的年龄（精确到月份）
SELECT 
	YEAR(CURDATE()) - YEAR(s_birth)
FROM student;
SELECT
	s_id,
	datediff(CURDATE(), s_birth) / 240
FROM student;


-- 47、查询本月过生日的学生（无法使用week、date(now()）
SELECT *
FROM student
WHERE MONTH(s_birth) = MONTH(CURDATE());