2023/04/16更新，此專欄50題有很多不嚴謹的地方且不便於測試，建議去牛客網或者LeetCode刷題，有測試用例且更貼近實際業務應用。
50題出自知乎專欄：
-- 1.查詢課程編號為“01”的課程比“02”的課程成績高的所有學生的學號（重點）
select a.s_id from (select s_id, c_id, s_score as score01 from score where c_id = "01") a
INNER join (select s_id, c_id, s_score as score02 from score where c_id = "02") b
on a.s_id = b.s_id
where score01 > score02;
-- 2、查詢平均成績大於60分的學生的學號和平均成績（簡單，第二道重點）
select s_id, avg(s_score) as avgs from score
group by s_id
having avgs > 60;
-- 3、查詢所有學生的學號、姓名、選課數、總成績（不重要）
select s_id, count(c_id) as 選課數, sum(s_score) as 總成績 from score GROUP BY s_id;
select s1.s_id as 學號, s_name as 姓名, count(c_id) as 選課數, sum(s_score) as 總成績 from student s1
inner join score s2
on s1.s_id = s2.s_id
group by s2.s_id;
-- 4、查詢姓“猴”的老師的個數（不重要）
SELECT count(t_id) from teacher where t_name like "猴%";
-- 5、查詢沒學過“張三”老師課的學生的學號、姓名（重點）
-- 學過張三課程的學生id
select s1.s_id from score s1
INNER join course c on s1.c_id = c.c_id
INNER join teacher t on t.t_id = c.t_id
where t_name = "張三";
select * from student where s_id not in (
select s1.s_id from score s1
INNER join course c on s1.c_id = c.c_id
INNER join teacher t on t.t_id = c.t_id
where t_name = "張三"
);
-- 6、查詢學過“張三”老師所教的所有課的同學的學號、姓名（重點）
-- 20230416更新，經評論提醒更新了解答，先選擇學過張三老師的課的學生學號，通過group by，找到學號出現次數等於張三老師所教課程數的學號。
select s1.s_id, s2.s_name from score s1
INNER JOIN student s2 on s1.s_id = s2.s_id
INNER join course c on s1.c_id = c.c_id
INNER join teacher t on t.t_id = c.t_id
WHERE t.t_id in (
SELECT t_id FROM teacher
WHERE teacher.t_name IN ("張三")
)
GROUP BY s1.s_id
HAVING COUNT(s1.s_id) = (
SELECT COUNT(c_id) FROM teacher t
INNER join course c on t.t_id = c.c_id
WHERE t.t_name IN ("張三")
);
-- 原解答
select * from student where s_id in (
select s1.s_id from score s1
INNER join course c on s1.c_id = c.c_id
INNER join teacher t on t.t_id = c.t_id
where t_name = "張三"
);
-- 7、查詢學過編號為“01”的課程並且也學過編號為“02”的課程的學生的學號、姓名（重點）
select s1.s_id, s_name from score s1
INNER join score s2 on s1.s_id = s2.s_id
INNER join student s on s1.s_id = s.s_id
where s1.c_id = "01" and s2.c_id = "02";
-- 8、查詢課程編號為“02”的總成績（不重點）
select sum(s_score) from score where c_id = "02";
-- 9、查詢所有課程成績小於60分的學生的學號、姓名
-- 考慮 所有課程成績中的最大值也小於60分
select score.s_id, s_name from score
INNER join student on score.s_id = student.s_id
GROUP BY s_id having max(s_score) < 60;
-- 10.查詢沒有學全所有課的學生的學號、姓名(重點)
-- 考慮 課程數小於課程總數的s_id
select s1.s_id, s_name from score s1
INNER join student s2
on s1.s_id = s2.s_id
group by s_id
HAVING count(c_id) < (select count(c_id) from course);
-- 11、查詢至少有一門課與學號為“01”的學生所學課程相同的學生的學號和姓名（重點）
select DISTINCT s1.s_id, s_name from student s1
INNER join score s2 on s1.s_id = s2.s_id
where c_id in
(select c_id from score where s_id = "01")
and s1.s_id != "01";
-- 12.查詢和“01”號同學所學課程完全相同的其他同學的學號(重點)
-- 需要確保該同學沒有學過“01”號同學沒學的課程，還需保證課程數相等
select s_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"
)
)
group by s_id
having count(*) = (
select count(*) from score where s_id = "01"
);
-- 13、查詢沒學過"張三"老師講授的任一門課程的學生姓名 和47題一樣（重點，能做出來）
select s_id, s_name from student where s_id not in (
select s.s_id from course c
INNER join score s on c.c_id = s.c_id
INNER join teacher t on c.t_id = t.t_id
where t.t_name = "張三"
);
-- 15、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（重點）
select s1.s_id, s2.s_name, avg(s_score) as 平均成績 from score s1
join student s2 on s1.s_id = s2.s_id
where s_score < 60
GROUP BY s_id
having count(s_score) >= 2;
-- 16、檢索"01"課程分數小於60，按分數降序排列的學生信息（和34題重復，不重點）
select st.* from student st
INNER join score sc on st.s_id = sc.s_id
where c_id = "01" and s_score < 60
ORDER BY s_score DESC;
-- 17、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(重重點與35一樣)
-- 有參考
select s_id "學號",
max(case when c_id = '01' then s_score else null end) as '語文',
max(case when c_id = '02' then s_score else null end) as '數學',
max(case when c_id = '03' then s_score else null end) as '英語',
avg(s_score) as 平均成績 from score GROUP BY s_id
order by 平均成績 desc;
-- 18.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 (超級重點)
select sc.c_id 課程ID, c_name 課程名稱, max(s_score) 最高分, min(s_score) 最低分, avg(s_score) 平均分,
avg(case when s_score >= 60 then 1.0 else 0.0 end) as 及格率,
avg(case when s_score >= 70 and s_score < 80 then 1.0 else 0.0 end) as 中等率,
avg(case when s_score >= 80 and s_score < 90 then 1.0 else 0.0 end) as 優良率,
avg(case when s_score >= 90 then 1.0 else 0.0 end) as 優秀率
from score sc
inner join course c on sc.c_id = c.c_id
group by sc.c_id;
-- 19、按各科成績進行排序，並顯示排名(重點row_number)
select *, ROW_NUMBER() over (ORDER BY s_score DESC) as 排名
from score;
-- 20、查詢學生的總成績並進行排名（不重點）
select s_id, sum(s_score) 總成績
from score
GROUP BY s_id
ORDER BY 總成績 DESC;
-- 21 、查詢不同老師所教不同課程平均分從高到低顯示(不重點)
select c.t_id, t.t_name, avg(s_score) 平均分
from course c
INNER join score s on c.c_id = s.c_id
INNER join teacher t on c.t_id = t.t_id
GROUP BY s.c_id
ORDER BY 平均分 DESC;
-- 22、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（重要 25類似）
select st.*, sc.s_score, ROW_NUMBER() over (PARTITION BY c_id ORDER BY s_score desc) as ranking
from score sc INNER join student st on sc.s_id = st.s_id;
-- 窗口函數限制不能直接使用ranking列？
select * from (
select st.*, sc.s_score, ROW_NUMBER() over (PARTITION BY c_id ORDER BY s_score desc) as ranking
from score sc INNER join student st on sc.s_id = st.s_id
) r
where ranking between 2 and 3;
-- 23、使用分段[100-85],[85-70],[70-60],[<60]來統計各科成績，分別統計:各分數段人數、課程ID和課程名稱(重點和18題類似)
select s.c_id, c.c_name,
sum(case when s_score <= 100 and s_score >= 85 then 1 else 0 end) "100-85",
sum(case when s_score < 85 and s_score >= 70 then 1 else 0 end) "85-70",
sum(case when s_score < 70 and s_score >= 60 then 1 else 0 end) "70-60",
sum(case when s_score < 60 then 1 else 0 end) "<60"
from score s
INNER join course c on s.c_id = c.c_id
group by s.c_id;
-- 24、查詢學生平均成績及其名次（同19題，重點）
select s_id, avg(s_score) avgs, ROW_NUMBER() over (order by avg(s_score) desc) ranking
from score group by s_id;
-- 25、查詢各科成績前三名的記錄（不考慮成績並列情況）（重點 與22題類似）
select c_id, st.s_name, s_score from
(select s_id, c_id, s_score, row_number() over (partition by c_id order by s_score desc) ranking
from score) s
INNER join student st on s.s_id = st.s_id
where ranking < 4;
-- 26、查詢每門課程被選修的學生數(不重點)
select c_id, count(c_id) from score
group by c_id;
-- 27、 查詢出只有兩門課程的全部學生的學號和姓名(不重點)
select st.s_id, st.s_name, count(c_id) from student st
INNER join score sc on st.s_id = sc.s_id
group by sc.s_id
having count(c_id) = 2;
-- 28、查詢男生、女生人數(不重點)
select s_sex, count(s_sex) from student group by s_sex;
-- 29 查詢名字中含有"風"字的學生信息（不重點）
select * from student
where s_name like "%風%";
-- 31、查詢1990年出生的學生名單（重點year）
select * from student
where year(s_birth) = 1990;
-- 32、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（不重要）
select st.s_id, st.s_name, avg(s_score) 平均成績 from student st
INNER join score sc on st.s_id = sc.s_id
GROUP BY st.s_id
having 平均成績 >= 85;
-- 33、查詢每門課程的平均成績，結果按平均成績升序排序，平均成績相同時，按課程號降序排列（不重要）
select c_id, avg(s_score) avgs from score
group by c_id
order by avgs asc, c_id desc;
-- 34、查詢課程名稱為"數學"，且分數低於60的學生姓名和分數（不重點）
select s_name, s_score from score sc
inner join student st on sc.s_id = st.s_id
inner join course c on sc.c_id = c.c_id
where c.c_name = "數學" and s_score < 60;
-- 35、查詢所有學生的課程及分數情況（重點）
select s_id,
max(case when c_name = '數學' then s_score else null end) 數學,
max(case when c_name = '語文' then s_score else null end) 語文,
max(case when c_name = '英語' then s_score else null end) 英語
from score s
INNER join course c on s.c_id = c.c_id
group by s_id;
-- 36、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（重點）
select st.s_id, s_name, c_name, s_score from student st
INNER join score sc on st.s_id = sc.s_id
inner join course c on sc.c_id = c.c_id
where s_score > 70;
-- 37、查詢不及格的課程並按課程號從大到小排列(不重點)
select st.s_id, s_name, c_name, sc.c_id, s_score from score sc
inner join student st on sc.s_id = st.s_id
inner join course c on sc.c_id = c.c_id
where s_score < 60
order by sc.c_id desc;
-- 38、查詢課程編號為03且課程成績在80分以上的學生的學號和姓名（不重要）
select sc.s_id, st.s_name
from score sc
inner join student st on sc.s_id = st.s_id
where sc.c_id = "03" and s_score > 80;
-- 39、求每門課程的學生人數（不重要）
select c_id, count(s_id) from score
group by c_id;
-- 40、查詢選修“張三”老師所授課程的學生中成績最高的學生姓名及其成績（重要top）
select s_name, s_score from student st
inner join score sc on st.s_id = sc.s_id
inner join course c on sc.c_id = c.c_id
inner join teacher t on c.t_id = t.t_id
where t.t_name = "張三"
order by s_score desc
limit 1;
-- 41.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 （重點）
-- 題意有點模糊，有參考
select distinct s1.s_id, s1.s_score, s1.c_id
from score s1
inner 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、查詢每門功課成績最好的前兩名（同22和25題）
select * from (
select s_id, c_id, s_score, ROW_NUMBER() over (PARTITION BY c_id order by s_score desc) ranking
from score) a
where ranking < 3;
-- 43、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（不重要）
select c_id, count(s_id) counts from score
group by c_id
having counts > 5
order by counts desc, c_id asc;
-- 44、檢索至少選修兩門課程的學生學號（不重要）
select s_id, count(c_id) from score
group by s_id
having count(c_id) >= 2;
-- 45、 查詢選修了全部課程的學生信息（重點划紅線地方）
select st.* from student st
inner join score sc on st.s_id = sc.s_id
group by s_id
having count(c_id) = (select count(*) from course);
-- 46、查詢各學生的年齡（精確到月份）
/* 時間戳 select from_unixtime(1645707893);
select unix_timestamp(now());*/
select s_name, timestampdiff(month, s_birth, now()) / 12 年齡
from student;
/*
47、48 這兩個周數問題很多，week(day)獲得的是day在該年度的第幾周，不能與其他年份的第幾周比較；
參考 select week("2020-01-05"), week("2018-01-05"); 結果一個為第一周，一個為第0周；
那么判斷生日所在周數需要將年份替換為當前日期的年份，再判斷該日期在now()對應年份的周數；
關於datediff差值小於7的判斷方法，完全不能保證兩個周數相同，比如周日和周一；
還需要考慮一年中最后一周的情況，這一系列都過於復雜，就不寫了。
*/
-- 47、查詢本周過生日的學生
-- 48、查詢下周過生日的學生
-- 49、查詢本月過生日的學生
select s_name, s_birth from student
where month(s_birth) = month(now());
-- 50、查詢下月過生日的學生
select s_name, s_birth from student
where month(s_birth) = (month("2020-12-01") + 1) % 12;