MySQL 50題練習
表名和欄位
–1.學生表
Student(s_id,s_name,s_birth,s_sex) –學生編號,學生姓名, 出生年月,學生性別
–2.課程表
Course(c_id,c_name,t_id) – –課程編號, 課程名稱, 教師編號
–3.教師表
Teacher(t_id,t_name) –教師編號,教師姓名
–4.成績表
Score(s_id,c_id,s_score) –學生編號,課程編號,分數
set sql_mode = ”;
set sql_mode = ‘NO_ENGINE_SUBSTITUTION,STRICT_TRANS_TABLES’;
測試數據
–建表
–學生表
CREATE TABLE Student(
s_id VARCHAR(20),
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)
);
–課程表
CREATE TABLE Course(
c_id VARCHAR(20),
c_name VARCHAR(20) NOT NULL DEFAULT ”,
t_id VARCHAR(20) NOT NULL,
PRIMARY KEY(c_id)
);
–教師表
CREATE TABLE Teacher(
t_id VARCHAR(20),
t_name VARCHAR(20) NOT NULL DEFAULT ”,
PRIMARY KEY(t_id)
);
–成績表
CREATE TABLE Score(
s_id VARCHAR(20),
c_id VARCHAR(20),
s_score INT(3),
PRIMARY KEY(s_id,c_id)
);
–插入學生表測試數據
insert into Student values(’01’ , ‘趙雷’ , ‘1990-01-01’ , ‘男’);
insert into Student values(’02’ , ‘錢電’ , ‘1990-12-21’ , ‘男’);
insert into Student values(’03’ , ‘孫風’ , ‘1990-05-20’ , ‘男’);
insert into Student values(’04’ , ‘李雲’ , ‘1990-08-06’ , ‘男’);
insert into Student values(’05’ , ‘周梅’ , ‘1991-12-01’ , ‘女’);
insert into Student values(’06’ , ‘吳蘭’ , ‘1992-03-01’ , ‘女’);
insert into Student values(’07’ , ‘鄭竹’ , ‘1989-07-01’ , ‘女’);
insert into Student values(’08’ , ‘王菊’ , ‘1990-01-20’ , ‘女’);
–課程表測試數據
insert into Course values(’01’ , ‘語文’ , ’02’);
insert into Course values(’02’ , ‘數學’ , ’01’);
insert into Course values(’03’ , ‘英語’ , ’03’);
–教師表測試數據
insert into Teacher values(’01’ , ‘張三’);
insert into Teacher values(’02’ , ‘李四’);
insert into Teacher values(’03’ , ‘王五’);
–成績表測試數據
insert into Score values(’01’ , ’01’ , 80);
insert into Score values(’01’ , ’02’ , 90);
insert into Score values(’01’ , ’03’ , 99);
insert into Score values(’02’ , ’01’ , 70);
insert into Score values(’02’ , ’02’ , 60);
insert into Score values(’02’ , ’03’ , 80);
insert into Score values(’03’ , ’01’ , 80);
insert into Score values(’03’ , ’02’ , 80);
insert into Score values(’03’ , ’03’ , 80);
insert into Score values(’04’ , ’01’ , 50);
insert into Score values(’04’ , ’02’ , 30);
insert into Score values(’04’ , ’03’ , 20);
insert into Score values(’05’ , ’01’ , 76);
insert into Score values(’05’ , ’02’ , 87);
insert into Score values(’06’ , ’01’ , 31);
insert into Score values(’06’ , ’03’ , 34);
insert into Score values(’07’ , ’02’ , 89);
insert into Score values(’07’ , ’03’ , 98);
MySQL50題
1、查詢”01″課程比”02″課程成績高的學生的資訊及課程分數
解法1：分別把課程01的成績和課程02的成績查詢出來之後再比較
SELECT a.*, d.01_score, d.02_score FROM student a INNER JOIN
(SELECT b.s_id, b.01_score, c.02_score FROM
(SELECT s_id, s_score as 01_score FROM score WHERE c_id=’01’)b INNER JOIN
(SELECT s_id, s_score as 02_score FROM score WHERE c_id=’02’)c ON b.s_id=c.s_id
WHERE b.01_score > c.02_score)d ON a.s_id=d.s_id;
解法2：全部查出來多重條件篩選
SELECT a.*, b.s_score as 01_score, c.s_score as 02_score FROM student a, score b, score c
WHERE a.s_id = c.s_id AND
a.s_id = b.s_id AND
b.c_id = ’01’ AND
c.c_id = ’02’ AND
b.s_score > c.s_score;
解法3：
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a
join score b on a.s_id=b.s_id and b.c_id=’01’
left join score c on a.s_id=c.s_id and c.c_id=’02’ or c.c_id = NULL where b.s_score>c.s_score;
2、查詢”01″課程比”02″課程成績低的學生的資訊及課程分數
同上題解法
3、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 a.s_name, ROUND(AVG(b.s_score), 2) as avg_score
FROM student a JOIN score b ON a.s_id=b.s_id
GROUP BY a.s_id
HAVING avg_score >= 60
ORDER BY avg_score DESC;
4、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 b.s_name, b.s_id, ROUND(SUM(a.s_score)/COUNT(a.c_id),2) as avg_score
FROM score a JOIN student b ON a.s_id=b.s_id
GROUP BY a.s_id HAVING avg_score < 60
UNION SELECT a.s_name, a.s_id, 0 as avg_score FROM student a WHERE a.s_id NOT in (SELECT DISTINCT s_id FROM score);
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 a.s_id, a.s_name, COUNT(b.c_id)as sum_course, SUM(b.s_score) as sum_score FROM student a JOIN score b on a.s_id=b.s_id GROUP BY a.s_id,a.s_name
UNION SELECT a.s_id, a.s_name, 0 as sum_sourse, 0 as sum_score FROM student a WHERE a.s_id NOT in (SELECT DISTINCT s_id FROM score);
6、查詢”李”姓老師的數量
SELECT COUNT(t_id) as ‘李姓老師數量’ FROM teacher WHERE t_name LIKE “李%” GROUP BY t_id;
7、查詢學過”張三”老師授課的同學的資訊
SELECT a.* FROM student a, score b, teacher c, course d
WHERE a.s_id=b.s_id AND
b.c_id = d.c_id AND
c.t_id = d.t_id AND
c.t_name = ‘張三’;
8、查詢沒學過”張三”老師授課的同學的資訊
SELECT a.* FROM student a WHERE a.s_id NOT in (SELECT a.s_id FROM student a, score b, teacher c, course d
WHERE a.s_id=b.s_id AND
b.c_id = d.c_id AND
c.t_id = d.t_id AND
c.t_name = ‘張三’);
9、查詢學過編號為”01″並且也學過編號為”02″的課程的同學的資訊
SELECT a.* FROM student a JOIN
(SELECT a.s_id FROM (SELECT s_id FROM score WHERE c_id=’01’)a INNER JOIN (SELECT s_id FROM score WHERE c_id=’02’)b on a.s_id=b.s_id)d on a.s_id=d.s_id;
10、查詢學過編號為”01″但是沒有學過編號為”02″的課程的同學的資訊
SELECT c.* FROM student c JOIN
(SELECT a.s_id FROM score a WHERE a.c_id=’01’ AND a.s_id NOT in (SELECT s_id FROM score WHERE c_id=’02’))d ON c.s_id=d.s_id;
11、查詢沒有學全所有課程的同學的資訊
SELECT a.*, COUNT(b.c_id) as sum_course FROM student a JOIN score b ON a.s_id=b.s_id GROUP BY a.s_id HAVING sum_course < 3;
12、查詢至少有一門課與學號為”01″的同學所學相同的同學的資訊
SELECT DISTINCT a.* FROM student a JOIN score b ON a.s_id=b.s_id WHERE b.c_id in
(SELECT c.c_id FROM score c WHERE c.s_id=’01’);
13、查詢和”01″號的同學學習的課程完全相同的其他同學的資訊
GROUP_CONCAT(expr) 配合 GROUP BY 可以將同一個分組中產生的值連接起來
SELECT c.*, d.course FROM student c JOIN
(SELECT a.s_id, a.course FROM
(SELECT s_id, GROUP_CONCAT(c_id) as course FROM score GROUP BY s_id)a JOIN (SELECT s_id, GROUP_CONCAT(c_id) as course FROM score GROUP BY s_id HAVING s_id=’01’)b
ON a.course=b.course)d ON c.s_id=d.s_id WHERE c.s_id not in (1);
14、查詢各科成績最高分、最低分和平均分： 以如下形式顯示：
課程 ID，課程 name，最高分，最低分，平均分，及格率，中等率，優良率，優秀率
及格為>=60，中等為：70-80，優良為：80-90，優秀為：>=90 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 a.c_id, b.c_name, MAX(a.s_score) as max_score, MIN(a.s_score) as min_score, ROUND(AVG(a.s_score), 2) as avg_score, COUNT(a.s_id)as count_stu,
CONCAT(ROUND(SUM(CASE when a.s_score >= 60 THEN 1 ELSE 0 END)/COUNT(), 2)100, ‘%’) as ‘及格率’,
CONCAT(ROUND(SUM(CASE when a.s_score >= 70 and a.s_score < 80 THEN 1 ELSE 0 END)/COUNT(), 2)100, ‘%’) as ‘中等率’,
CONCAT(ROUND(SUM(CASE when a.s_score >= 80 and a.s_score < 80 THEN 1 ELSE 0 END)/COUNT(),2)100, ‘%’) as ‘優良率’,
CONCAT(ROUND(SUM(CASE when a.s_score >= 90 THEN 1 ELSE 0 END)/COUNT(),2)100, ‘%’) as ‘優秀率’
FROM score a JOIN course b ON a.c_id=b.c_id GROUP BY a.c_id ORDER BY count_stu DESC, a.c_id ASC;
15、按各科成績進行排序，並顯示排名， Score 重複時保留名次空缺
15.1 按各科成績進行排序，並顯示排名， Score 重複時合併名次
select a.s_id,a.c_id,
@i:=@i +1 as i, #保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank, #不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s;
16、查詢學生的總成績，並進行排名，總分重複時保留名次空缺
16.1 查詢學生的總成績，並進行排名，總分重複時不保留名次空缺
SELECT a.s_id, a.sum_score,
@i:=@i+1 as i, # 保留排名
@k:=(CASE WHEN @score=a.sum_score THEN @k ELSE @i END) as rank, # 不保留排名
@score:=a.sum_score as score FROM
(SELECT s_id, sum(s_score)as sum_score FROM score GROUP BY s_id ORDER BY sum_score DESC)a, (SELECT @k:=0, @i:=0, @score:=0)s;
17、統計各科成績各分數段人數：課程編號，課程名稱，[100-85]，[85-70]，[70-60]，[60-0] 及所佔百分比
SELECT score.c_id, c_name,
CONCAT(ROUND(sum(CASE WHEN s_score <=100 and s_score>85 THEN 1 ELSE 0 END)/COUNT(),2)100,’%’) as ‘[100-85]’,
CONCAT(ROUND(sum(CASE WHEN s_score <=85 and s_score>70 THEN 1 ELSE 0 END)/COUNT(),2)100,’%’) as ‘[85-70]’,
CONCAT(ROUND(sum(CASE WHEN s_score <=70 and s_score>60 THEN 1 ELSE 0 END)/COUNT(),2)100,’%’) as ‘[70-60]’,
CONCAT(ROUND(sum(CASE WHEN s_score <=60 and s_score>=0 THEN 1 ELSE 0 END)/COUNT(),2)100,’%’) as ‘[60-0]’
FROM score JOIN course ON score.c_id=course.c_id GROUP BY score.c_id;
18、查詢各科成績前三名的記錄
(SELECT c_id, s_score FROM score WHERE c_id = ’01’ ORDER BY s_score DESC LIMIT 3)
UNION (SELECT c_id, s_score FROM score WHERE c_id = ’02’ ORDER BY s_score DESC LIMIT 3)
UNION (SELECT c_id, s_score FROM score WHERE c_id = ’03’ ORDER BY s_score DESC LIMIT 3);
19、查詢每門課程被選修的學生數
SELECT c_id, COUNT(s_id)as sum_stu FROM score GROUP BY c_id;
20、查詢出只選修兩門課程的學生學號和姓名
SELECT a.s_id, b.s_name, COUNT(a.c_id)as sum_course FROM score a JOIN student b ON a.s_id=b.s_id GROUP BY a.s_id HAVING sum_course = 2;
21、查詢男生、女生人數
SELECT s_sex, COUNT(s_id)as sex_count FROM student GROUP BY s_sex;
22、查詢名字中含有「風」字的學生資訊
SELECT * FROM student WHERE s_name LIKE ‘%風%’;
23、查詢同名同性學生名單，並統計同名人數
SELECT s_name,s_sex, count()as ‘同名人數’
FROM student
GROUP BY s_name,s_sex
HAVING COUNT() >1;
24、查詢 1990 年出生的學生名單
SELECT * FROM student WHERE YEAR(s_birth)=’1990′;
25、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(s_score),2)as avg_score FROM score GROUP BY c_id ORDER BY avg_score DESC, c_id;
26、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(SUM(a.s_score)/COUNT(a.c_id), 2) as avg_score FROM score a JOIN student b on a.s_id=b.s_id GROUP BY a.s_id ORDER BY avg_score DESC
27、查詢課程名稱為「數學」，且分數低於 60 的學生姓名和分數
SELECT a.s_id, b.c_name,c.s_name, a.s_score FROM score a JOIN course b on a.c_id=b.c_id JOIN student c ON a.s_id=c.s_id WHERE b.c_name=’數學’ AND a.s_score < 60;
28、查詢所有學生的課程及分數情況（存在學生沒成績，沒選課的情況）
SELECT * FROM score
29、查詢任何一門課程成績在 70 分以上的學生姓名、課程名稱和分數
取反，先把所有課程成績在70分以下的學生id找出來(總課程計數和70以下課程計數一樣的就是全部課程成績在70分以下的)，取反就是任何一門課程成績在70分以上的同學
SELECT d.s_name, f.c_name, e.s_score FROM student d JOIN score e ON d.s_id=e.s_id JOIN course f ON e.c_id = f.c_id WHERE d.s_id NOT in
(SELECT a.s_id FROM
(SELECT s_id, COUNT(c_id)as course_count FROM score GROUP BY s_id)a JOIN
(SELECT s_id, COUNT(c_id)as course_count FROM score WHERE s_score < 70 GROUP BY s_id)b ON (a.course_count=b.course_count and a.s_id=b.s_id));
30、查詢不及格的課程
SELECT a.s_id,b.s_name, c.c_name, a.s_score FROM score a JOIN student b ON a.s_id=b.s_id JOIN course c ON a.c_id=c.c_id WHERE s_score < 60;
31、查詢課程編號為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, a.c_id,a.s_score FROM score a JOIN student b ON a.s_id=b.s_id WHERE a.c_id=’01’ AND a.s_score >= 80;
32、求每門課程的學生人數
SELECT c_id, COUNT(s_id) as count_stu FROM score GROUP BY c_id;
33、查詢選修”張三”老師所授課程的學生中，成績最高的學生資訊及其成績
SELECT d.*, b.c_name, c.s_score FROM teacher a
JOIN course b ON a.t_id=b.t_id
JOIN score c ON b.c_id=c.c_id
JOIN student d ON c.s_id=d.s_id
WHERE a.t_name=’張三’ ORDER BY c.s_score DESC LIMIT 1;
select a.*,b.s_score,b.c_id,c.c_name 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
where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name=’張三’)
and b.s_score in (select MAX(s_score) from score where c_id=’02’)
34、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 DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score;
35、查詢每門功成績最好的前兩名
查詢兩張score表，找出b表中比a表中單科成績高的計數（雙循環比較 O(n^2)），小於等於2那就是找到了前兩位
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id;
36、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(s_id)as stu_count FROM score GROUP BY c_id HAVING stu_count > 5 ORDER BY stu_count DESC, c_id;
37、檢索至少選修兩門課程的學生學號
SELECT s_id, COUNT(c_id)as course_count FROM score GROUP BY s_id HAVING course_count >= 2;
38、查詢選修了全部課程的學生資訊
SELECT b.*,COUNT(a.c_id) as course_count FROM score a JOIN student b ON a.s_id=b.s_id GROUP BY a.s_id HAVING course_count=(SELECT COUNT(c_id) FROM course);
39、查詢各學生的年齡
-- 按照出生日期來算，當前月日 < 出生年月的月日則，年齡減一
