【Mysql學習之旅-2】經典sql面試題及答案分析

前言

在學習了Mysql的基礎知識後，我們用一套sql練習題來實戰一下。

關於練習所需要的數據，讓我們繼續玩壞倚天屠龍的江湖。

1、學生表 student（s_id：學生id，s_name：學生姓名，s_birth：學生生日，s_sex：學生性別）：

2、教師表teacher（t_id：教師id，t_name：教師姓名)

3、課程表 course（c_id：課程id，c_name：課程名稱，t_id：教師id）：

4、成績表 score（s_id：學生id，c_id：課程id，score：分數）

初始化數據的sql附在文章末尾

題目開始，先上幾道硬菜，磕到牙齒的可以移步進階題目和基礎題目區壓壓驚。

挑戰

1、查詢各科成績前三名的記錄

--方法1：
SELECT 
	s_name,a.s_id,a.c_id,a.score 
FROM 
	score a 
LEFT JOIN 
	score b 
ON 
	a.c_id = b.c_id AND a.score<b.score
JOIN 
	student
ON
	a.s_id=student.s_id
GROUP BY 
	a.s_id,a.c_id,a.score HAVING COUNT(b.s_id)<3
ORDER BY 
	a.c_id,a.score DESC

--方法2：
SELECT 
	a.s_id,a.c_id,a.score 
FROM 
	score a
WHERE 
	(SELECT COUNT(1) FROM score b WHERE b.c_id=a.c_id 
	AND b.score>=a.score)<=2 ORDER BY a.c_id

簡評：非常難以理解的一道題目，即使想透徹了也不容易記住，堪稱勸退型難題。面試中，偶爾會考察。
難度：⭐⭐⭐⭐⭐
易考率：⭐⭐
思路：
1、SELECT * FROM score a JOIN score b ON a.c_id=b.c_id WHERE a.c_id=‘03’ 這一句，形成了03課程的所有學員的兩兩組合（帶上03，是爲了簡化數據，便於分析）

2、SELECT * FROM score a JOIN score b ON a.c_id=b.c_id WHERE a.c_id=‘03’ AND a.score<b.score又篩選出每個學員03課程比其他人低的情況（某個學員分越低，記錄越多，沒記錄說明是最高分）

3、將這些記錄按s_id分組，篩選出記錄數小於3的（說明最多有兩個人比自己分數高），即是前3名

2、查詢“01”課程比“02”課程成績高的所有學生的學號

SELECT 
	*
FROM 
	(SELECT s_id,score,c_id FROM score WHERE c_id='01') a
JOIN
	(SELECT s_id,score,c_id FROM score WHERE c_id='02') b
ON 
	a.s_id=b.s_id 
WHERE 
	a.score>b.score

簡評：不容易想到，但答案相對第1題來說，就好理解多了。
難度：⭐⭐⭐⭐
頻率：⭐⭐
思路：通過SELECT s_id,score,c_id FROM score WHERE c_id='01’和SELECT s_id,score,c_id FROM score WHERE c_id='02’的兩個臨時表聯查，得到每個學員的01、02課程成績情況：

接下來，a.score>b.score條件直接篩選就可以了。

3、查詢和"04"號的同學學習的課程完全相同的其他同學的信息

SELECT 
	score.s_id,COUNT(DISTINCT c_id) 
FROM 
	student JOIN score ON student.s_id=score.s_id 
GROUP BY 
	score.s_id 
HAVING COUNT(DISTINCT c_id)= 
	(SELECT COUNT(c_id) FROM score WHERE s_id='04') 
AND 
	s_id 
NOT IN
	-- 篩選出學過（04號同學未學課程）的學生，通過not in排除掉
	(SELECT s_id FROM score WHERE c_id IN
	(SELECT DISTINCT(c_id) FROM course WHERE c_id 
	NOT IN
	(SELECT c_id FROM score WHERE s_id='04'))) 
AND score.s_id!='04'

簡評：難在思路的迂迴性。面試中，很多有難度的sql也大多難在直接去考慮，難以獲得答案。
難度：⭐⭐⭐⭐
頻率：⭐⭐
思路：通常思路，我們先獲取04號同學的課程，再遍歷其他同學，以此篩選。這種方式適用於編程，單純的sql實現不了，我們應該從下面這個思路去考慮------>我和04號同學學習的課程數目一樣多，且04號同學沒學過的我也沒學過，那麼我不就是和04號同學學習的課程一樣嗎？

4、按平均成績從高到低顯示所有學生的所有課程的成績以及平均成績

SELECT 
	a.s_id,b.s_name,
	MAX(CASE a.c_id WHEN '01' THEN a.score END ) 內功, 
	MAX(CASE a.c_id WHEN '02' THEN a.score END ) 劍法, 
	MAX(CASE a.c_id WHEN '03' THEN a.score END ) 拳法, 
	AVG(a.score) 
FROM 
	score a 
JOIN 
	student b ON a.s_id=b.s_id 
GROUP BY a.s_id ORDER BY AVG(a.score) DESC

簡評：思路不難，關鍵在於理解透徹case when語法。
難度：⭐⭐⭐
頻率：⭐⭐

SELECT a.c_id,b.c_name,MAX(score),MIN(score),ROUND(AVG(score),2),
	ROUND(100*(SUM(case when a.score>=60 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 及格率,
	ROUND(100*(SUM(case when a.score>=70 and a.score<=80 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 中等率,
	ROUND(100*(SUM(case when a.score>=80 and a.score<=90 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 優良率,
	ROUND(100*(SUM(case when a.score>=90 then 1 else 0 end)/SUM(case when a.score then 1 else 0 end)),2) as 優秀率
FROM 
	score a 
LEFT JOIN 
	course b 
ON a.c_id = b.c_id GROUP BY a.c_id,b.c_name

簡評：在case when語法的基礎上，進一步掌握sql語句中進行邏輯運算的技巧
難度：★★★☆
頻率：★★★

5、按各科成績進行排序，並顯示排名

SELECT a.s_id,a.c_id,
	@i:=@i + 1 AS 排名,
    @j:=(CASE WHEN @score=a.score THEN @j ELSE @i END) AS 並排排名,
    @score:=a.score AS score
FROM 
	(SELECT s_id,c_id,score FROM score GROUP BY s_id,c_id ORDER BY score DESC) a,
	(SELECT @i:= 0,@j:= 0,@score:= 0) s

簡評：Mysql中沒有rank函數，只能使用僞列的概念去實現排名算法。
難度：★★★★
頻率：★★
思路：
1、SELECT s_id,c_id,score FROM score GROUP BY s_id,c_id ORDER BY score DESC先進行了排序
2、(SELECT @i:=0,@j:=0,@score:=0) s對@i、@j、@score進行了初始化，初始值都爲0
3、@i:=@i + 1，遍歷每一行，@i逐行自增
4、@j:=(CASE WHEN @score=a.score THEN @j ELSE @i END)，遍歷每一行，邏輯判斷，@score與上一行的score相等則@j=@i，否則則自增一次

6、查詢學生的總成績並進行排名

SELECT a.s_id,
	@i:=@i+1 as i,
	@j:=(CASE WHEN @score=a.sum_score THEN @j ELSE @i END) AS rank,
	@score:=a.sum_score AS score
FROM 
	(SELECT s_id,SUM(score) AS sum_score FROM score GROUP BY s_id ORDER BY sum_score DESC)a,
	(SELECT @i:=0,@j:=0,@score:=0) s

簡評：與上一題基本差不多，不同的是a這個臨時表的數據內容而已。
難度：★★★★
頻率：★★

7、查詢所有課程的成績第2名到第3名的學生信息及該課程成績

SELECT 
	student.*,c.rank,c.score,c.c_id 
FROM 
	(SELECT a.s_id,a.score,a.c_id,@i:=@i+1 as rank from score a,(SELECT @i:=0)s 
	WHERE a.c_id='01' ORDER BY a.score DESC) c
LEFT JOIN 
	student ON c.s_id=student.s_id
WHERE rank BETWEEN 2 AND 3
UNION ALL 
SELECT 
	student.*,c.rank,c.score,c.c_id 
FROM 
	(SELECT a.s_id,a.score,a.c_id,@j:=@j+1 as rank from score a,(SELECT @j:=0)s 
	WHERE a.c_id='02' ORDER BY a.score DESC) c
LEFT JOIN 
	student ON c.s_id=student.s_id
WHERE 
	rank BETWEEN 2 AND 3
UNION ALL 
SELECT 
	student.*,c.rank,c.score,c.c_id 
FROM 
	(SELECT a.s_id,a.score,a.c_id,@k:=@k+1 as rank from score a,(SELECT @k:=0)s 
	WHERE a.c_id='03'  ORDER BY a.score DESC) c
LEFT JOIN 
	student ON c.s_id=student.s_id
WHERE rank BETWEEN 2 AND 3

簡評：sql很長，不要被嚇到，其實是3個相同的部分（where條件不同）通過UNION ALL合併而已。另外，篩選2、3名也是在排名的基礎上進一步篩選。
難度：★★★★☆
頻率：★★
思路：
1、先將01課程的所有人的分數排序，再進一步根據rank BETWEEN 2 AND 3篩選出2、3名
2、使用同樣方法，篩選02、03課程數據，使用 UNION ALL合併查詢結果

8、統計各科成績各分數段人數：課程編號、課程名稱、[80-100]，[60-80]，[0-60]及所佔百分比

SELECT 
	DISTINCT e.c_name,a.c_id,b.`80-100`,b.百分比,c.`60-80`,c.百分比,d.`0-60`,d.百分比 
FROM 
	score a
LEFT JOIN 
	(SELECT c_id,SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END) AS `80-100`,
	ROUND(100*(SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 end)/count(*)),2)
	AS 百分比 FROM score GROUP BY c_id)b ON a.c_id=b.c_id
LEFT JOIN 
	(SELECT c_id,SUM(CASE WHEN score >60 AND score <=80 THEN 1 ELSE 0 END) AS `60-80`,
	ROUND(100*(SUM(CASE WHEN score >60 AND score <=80 THEN 1 ELSE 0 end)/count(*)),2) 
	AS 百分比 FROM score GROUP BY c_id)c ON a.c_id=c.c_id
LEFT JOIN 
	(SELECT c_id,SUM(CASE WHEN score >=0 AND score <=60 THEN 1 ELSE 0 END) AS `0-60`,
	ROUND(100*(SUM(CASE WHEN score >=0 AND score <=60 THEN 1 ELSE 0 END)/count(*)),2) 
	AS 百分比 FROM score GROUP BY c_id)d ON a.c_id=d.c_id
LEFT JOIN 
	course e ON a.c_id = e.c_id

簡評：CASE WHEN和sql語句內進行邏輯運算的又一次應用。
難度：★★★★
頻率：★★
思路：SELECT c_id,SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END) AS80-100, ROUND(100*(SUM(CASE WHEN score >80 AND score <=100 THEN 1 ELSE 0 END)/count(*)),2) AS 百分比 FROM score GROUP BY c_id得到所有課程80-100分成績的統計臨時表：

然後再用同樣的方式，得到0-60、60-80的統計臨時表，三表聯查。

9、查詢學生平均成績及其名次

SELECT a.s_id,
	@i:=@i+1 as '不保留空缺排名',
	@k:=(CASE WHEN @avg_score=a.avg_s THEN @k ELSE @i END) AS '保留空缺排名',
	@avg_score:=avg_s AS '平均分'
FROM 
	(SELECT s_id,ROUND(AVG(score),2) AS avg_s FROM 
	score GROUP BY s_id ORDER BY avg_s DESC)a,
	(SELECT @avg_score:=0,@i:=0,@k:=0)b;

簡評：與第5題類似，沒什麼特別需要說的。
難度：★★★★
頻率：★★

10、查詢各學生的年齡

SELECT 
	s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - 
	(CASE WHEN DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') THEN 0 ELSE 1 END)) 
	AS age
FROM student;

簡評：本身的實現思路並不難，DATE_FORMAT掌握其用法即可。

11、分別查詢本週、下週、本月、下月過生日的學生

--本週
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
--下週
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 = WEEK(s_birth)
--本月
SELECT * FROM student WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
--下月
SELECT * FROM student WHERE MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

進階

1、查詢及格學生（每門課程的分數>=60）的學生姓名

SELECT s_name FROM 
	student 
WHERE s_id NOT IN
	(SELECT DISTINCT(s_id) FROM score WHERE score<60)

或者：

SELECT s_name FROM 
	student a JOIN score b ON a.s_id=b.s_id 
GROUP BY b.s_id HAVING MIN(score)>60

2、查詢平均成績小於60分的同學的學生編號和學生姓名和平均成績 (包括有成績的和無成績的)

SELECT 
	s_name,ROUND(avg(score),2) AS avg 
FROM 
	student 
JOIN 
	score ON student.s_id=score.s_id 
GROUP BY 
	score.s_id HAVING avg<60 
UNION ALL 
SELECT 
	s_name,0 AS avg 
FROM 
	student 
WHERE s_id NOT IN(SELECT DISTINCT(s_id) FROM score)

-- 方法2
SELECT 
	s_name,IFNULL(ROUND(avg(score),2),0) AS avg 
FROM 
	student 
LEFT JOIN 
	score ON student.s_id=score.s_id 
GROUP BY score.s_id HAVING avg<60 OR avg is NULL

3、查詢學過"張三丰"老師授課的同學的信息

SELECT 
	* 
FROM 
	score JOIN student ON score.s_id=student.s_id 
WHERE c_id IN
	(SELECT c_id FROM course c JOIN teacher t ON c.t_id=t.t_id WHERE t_name='張三丰')

4、查詢沒學過"張三丰"老師授課的同學的信息

SELECT 
	* 
FROM 
	student 
WHERE s_id NOT IN
	(SELECT s_id FROM score WHERE c_id 
IN
    (SELECT c_id FROM course c JOIN teacher t ON c.t_id=t.t_id WHERE t_name='張三丰'))

5、查詢沒有學全所有課程的同學的信息

SELECT 
	s_name,COUNT(c_id) AS count 
FROM 
	student 
LEFT JOIN 
	score ON student.s_id=score.s_id 
GROUP BY 
	student.s_id HAVING count<(SELECT COUNT(1) FROM course);

6、查詢至少有一門課與學號爲"01"的同學所學相同的同學的信息

SELECT * FROM 
	student 
WHERE s_id IN
	(SELECT DISTINCT a.s_id FROM score a WHERE a.c_id 
	IN
	(SELECT a.c_id FROM score a WHERE a.s_id='01'));

7、查詢兩門及其以上不及格課程的同學的學號，姓名及其平均成績

SELECT 
	a.s_id,a.s_name,ROUND(AVG(b.score)) 
FROM  
	student a 
LEFT JOIN 
	score b ON a.s_id = b.s_id
WHERE a.s_id IN
	(SELECT s_id from score WHERE score<60 GROUP BY s_id HAVING count(1)>=2)
GROUP BY a.s_id,a.s_name

8、查詢所有學生的課程及分數情況；

SELECT 
	a.s_id,a.s_name,
	SUM(CASE c.c_name WHEN '內功' THEN b.score ELSE 0 END) AS '內功',
	SUM(CASE c.c_name WHEN '劍法' THEN b.score ELSE 0 END) AS '劍法',
	SUM(CASE c.c_name WHEN '拳法' THEN b.score ELSE 0 END) AS '拳法',
	SUM(b.score) as  '總分'
FROM 
	student a 
LEFT JOIN 
	score b ON a.s_id = b.s_id 
LEFT JOIN 
	course c ON b.c_id = c.c_id 
GROUP BY a.s_id,a.s_name

9、查詢選修"張三丰"老師所授課程的學生中，成績最高的學生信息及其成績

SELECT 
	s_id,MAX(score) 
FROM 
	score 
WHERE 
	c_id 
IN
	(SELECT c_id FROM teacher JOIN course 
	ON teacher.t_id=course.t_id WHERE t_name='張三丰')

10、查詢不同課程成績相同的學生的學生編號、課程編號、學生成績

SELECT * FROM score a,score b where a.c_id != b.c_id and a.score = b.score

11、統計每門課程的學生選修人數（超過5人的課程才統計）。要求輸出課程號和選修人數，查詢結果按人數降序排列，若人數相同，按課程號升序排列

SELECT 
	c_id,COUNT(*) AS total 
FROM 
	score 
GROUP BY c_id HAVING total>4 ORDER BY total DESC,c_id ASC

12、查詢選修了全部課程的學生信息

SELECT 
	* FROM student WHERE s_id 
IN
	(SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*)=(SELECT COUNT(*) FROM course))

基礎

1、查詢不及格的課程

SELECT 
	a.s_id,a.c_id,b.c_name,a.score 
FROM 
	score a 
LEFT JOIN 
	course b ON a.c_id = b.c_id
WHERE a.score<60

2、查詢課程編號爲01且課程成績在80分以上的學生的學號和姓名

SELECT 
	a.s_id,b.s_name 
FROM 
	score a 
LEFT JOIN 
	student b ON a.s_id = b.s_id
WHERE a.c_id = '01'	AND a.score>80

3、查詢每個同學的姓名、選課數、總成績

SELECT 
	s_name,COUNT(c_id),SUM(score) 
FROM 
	student a 
LEFT JOIN 
	score b ON a.s_id=b.s_id 
GROUP BY b.s_id

4、查詢平均成績大於等於60分的同學的學生編號和學生姓名和平均成績

SELECT 
	s_name,ROUND(avg(score),2) AS avg 
FROM 
	student 
JOIN 
	score ON student.s_id=score.s_id 
GROUP BY score.s_id HAVING avg>60

5、查詢所有同學的學生編號、學生姓名、選課總數、所有課程的總成績

SELECT student.s_name,COUNT(c_id),IFNULL(SUM(score),0) AS sum FROM student LEFT JOIN score ON student.s_id=score.s_id GROUP BY student.s_id ORDER BY sum DESC

6、查詢"張"姓老師的數量

SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '張%'

7、查詢"01"課程分數小於60，按分數降序排列的學生信息

SELECT 
	a.*,b.c_id,b.score
FROM 
	student a,score b
WHERE 
	a.s_id=b.s_id 
AND
	b.c_id='01' AND b.score<60
ORDER BY
	b.score DESC;

8、查詢任何一門課程成績在70分以上的姓名、課程名稱和分數

SELECT 
	a.s_name,b.c_name,c.score 
FROM 
	course b 
LEFT JOIN 
	score c ON b.c_id = c.c_id
LEFT JOIN 
	student a ON a.s_id=c.s_id 
WHERE c.score>=70

9、查詢學過編號爲"01"並且也學過編號爲"02"的課程的同學的信息

SELECT a.* FROM 
	student a,score b,score c 
WHERE a.s_id = b.s_id AND a.s_id = c.s_id AND b.c_id='01' AND c.c_id='02';

10、查詢學過編號爲"01"但是沒有學過編號爲"02"的課程的同學的信息

SELECT a.* FROM 
	student a 
WHERE 
	a.s_id 
IN 
	(SELECT s_id FROM score WHERE c_id='01' ) 
AND 
	a.s_id NOT IN(SELECT s_id FROM score WHERE c_id='02')

11、查詢男生、女生人數

SELECT s_sex,COUNT(s_sex) FROM student GROUP BY s_sex

12、查詢名字中含有"圓"字的學生信息

SELECT * FROM student WHERE s_name LIKE '%圓%';

13、查詢同名同性學生名單，並統計同名人數

SELECT 
	a.s_name,a.s_sex,count(*) 
FROM 
	student a JOIN student b 
ON 
	a.s_id !=b.s_id and a.s_name = b.s_name 
AND 
	a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex

14、查詢1990年出生的學生名單

SELECT s_name FROM student WHERE s_birth LIKE '1337%'

15、查詢每門課程的平均成績，結果按平均成績降序排列，平均成績相同時，按課程編號升序排列

SELECT 
	c_id,ROUND(AVG(score),2) AS avg_score 
FROM 
	score 
GROUP BY c_id ORDER BY avg_score DESC,c_id ASC

16、查詢平均成績大於等於85的所有學生的學號、姓名和平均成績

SELECT 
	a.s_id,b.s_name,ROUND(avg(a.score),2) AS avg 
FROM 
	score a
LEFT JOIN 
	student b on a.s_id=b.s_id 
GROUP BY s_id HAVING avg>=85

17、查詢課程名稱爲"內功"，且分數低於60的學生姓名和分數

SELECT 
	a.s_name,b.score 
FROM 
	score b 
JOIN 
	student a ON a.s_id=b.s_id 
WHERE 
	b.c_id=(SELECT c_id FROM course WHERE c_name ='內功') 
AND b.score<60

18、求每門課程的學生人數

SELECT c_id,count(*) FROM score GROUP BY c_id;

19、檢索至少選修兩門課程的學生學號

SELECT s_id,count(*) AS count FROM score GROUP BY s_id HAVING count>=2

數據初始化

-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `s_id` varchar(20) NOT NULL,
  `s_name` varchar(20) NOT NULL DEFAULT '',
  `s_birth` varchar(20) NOT NULL DEFAULT '',
  `s_sex` varchar(10) NOT NULL DEFAULT '',
  PRIMARY KEY (`s_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES ('01', '張無忌', '1337-06-17', '男');
INSERT INTO `student` VALUES ('02', '宋青書', '1333-12-21', '男');
INSERT INTO `student` VALUES ('03', '丁敏君', '1336-06-21', '女');
INSERT INTO `student` VALUES ('04', '周芷若', '1340-09-26', '女');
INSERT INTO `student` VALUES ('05', '圓真', '1320-12-01', '男');
INSERT INTO `student` VALUES ('06', '圓覺', '1316-03-01', '男');
INSERT INTO `student` VALUES ('07', '殷梨亭', '1310-07-01', '男');
INSERT INTO `student` VALUES ('08', '紀曉芙', '1310-01-20', '女');


-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `c_id` varchar(20) NOT NULL,
  `c_name` varchar(20) NOT NULL DEFAULT '',
  `t_id` varchar(20) NOT NULL,
  PRIMARY KEY (`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES ('01', '內功', '03');
INSERT INTO `course` VALUES ('02', '劍法', '02');
INSERT INTO `course` VALUES ('03', '拳法', '01');

-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `t_id` varchar(20) NOT NULL,
  `t_name` varchar(20) NOT NULL DEFAULT '',
  PRIMARY KEY (`t_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '張三丰');
INSERT INTO `teacher` VALUES ('02', '滅絕師太');
INSERT INTO `teacher` VALUES ('03', '空見');

-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
  `s_id` varchar(20) NOT NULL,
  `c_id` varchar(20) NOT NULL,
  `score` int(3) DEFAULT NULL,
  PRIMARY KEY (`s_id`,`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of score
-- ----------------------------
INSERT INTO `score` VALUES ('01', '01', '99');
INSERT INTO `score` VALUES ('01', '02', '96');
INSERT INTO `score` VALUES ('01', '03', '90');
INSERT INTO `score` VALUES ('02', '01', '59');
INSERT INTO `score` VALUES ('02', '02', '75');
INSERT INTO `score` VALUES ('02', '03', '68');
INSERT INTO `score` VALUES ('03', '01', '55');
INSERT INTO `score` VALUES ('03', '02', '50');
INSERT INTO `score` VALUES ('03', '03', '60');
INSERT INTO `score` VALUES ('04', '01', '77');
INSERT INTO `score` VALUES ('04', '02', '88');
INSERT INTO `score` VALUES ('05', '01', '88');
INSERT INTO `score` VALUES ('05', '02', '90');
INSERT INTO `score` VALUES ('06', '01', '86');
INSERT INTO `score` VALUES ('06', '03', '99');
INSERT INTO `score` VALUES ('07', '02', '58');
INSERT INTO `score` VALUES ('07', '03', '98');