sql習題及答案

 

sql習題：http://www.cnblogs.com/wupeiqi/articles/5729934.html

習題答案參考：https://www.cnblogs.com/wupeiqi/articles/5748496.html  （有些答案有錯）

 

 

 

-- SELECT count(*) from score WHERE num>60; -- 查找分數大於60的個數
-- select count(cid),teacher_id from course group by teacher_id;
-- select tid,teacher.tname,course.cname from course left join teacher on course.teacher_id = teacher.tid;
-- select count(sid),gender from student GROUP BY gender  -- 男女生個數

-- 二、查詢「生物」課程比「物理」課程成績高的全部學生的學號；
        -- 找出生物成績，找出物理成績，聯合這兩張臨時表，找出B.num > P.num
-- select B.student_id,B.cname,B.num as b_num,P.cname,P.num as p_num from 
-- (select * from score left join course on score.course_id = course.cid where cname = '生物') as B
-- inner join
-- (select * from score left join course on score.course_id = course.cid where cname = '物理') as P
-- on B.student_id = P.student_id where B.num > P.num;;
-- 三、查詢平均成績大於60分的同窗的學號和平均成績;(進階：以及姓名)
            -- 首先選擇出平均分大於60分的同窗的學號，再和學生表join，選擇出姓名
-- SELECT
--     student.sid,
--     student.sname,
--     b1.avg_num 
-- FROM
--     ( SELECT avg( num ) AS avg_num, student_id FROM score GROUP BY student_id HAVING avg( num ) > 60 ) AS b1
--     LEFT JOIN student ON b1.student_id = student.sid;

-- 四、查詢全部同窗的學號、姓名、選課數、總成績；（兩種解法，另外一種是先把表連起來再group by）
                -- 這裏注意 count(1)的用法， 相似select age,1 from t1; 會出現列名1，屬性全爲1
            -- 首先成績表和學生表連表，再根據學號進行分組，而後選擇出學號，姓名，聚合學科數，求和num
-- SELECT
--     student.sid,
--     student.sname,
--     b2.course_num,
--     b2.sum_num 
-- FROM
--     ( SELECT student_id, count( course_id ) AS course_num, sum( num ) AS sum_num FROM score GROUP BY student_id ) AS b2
--     LEFT JOIN student ON b2.student_id = student.sid;

-- select student.sid,student.sname,count(1) as course_num,sum(num)  from score left join student on score.student_id = student.sid group by score.student_id 

-- 五、查詢姓「李」的老師的個數；
-- select count(1) from teacher where tname like '李%';

-- 六、查詢沒學過「李平」老師課的同窗的學號、姓名；
            -- 首先課程表和老師表join，選擇出李平老師的課程id，而後在成績表中把選擇了 這些課程id的學號選出，用學號分組後，用not in 從學生表中，選擇出沒有選擇過課程的學號和姓名
-- select sid,sname from student where sid not in (select student_id from score where course_id in (select course.cid from course left join teacher on course.teacher_id=teacher.tid where tname = '李平老師') group by student_id    );
-- 

-- 七、查詢學過「1」課程而且也學過編號「2」課程的同窗的學號、姓名；
--                 先查到既選擇001又選擇002課程的全部同窗
--         根據學生進行分組，若是學生數量等於2表示，兩門均已選擇
-- select student.sid,student.sname from
-- (select student_id from score where course_id in (1,2) group by student_id having count(*)>1) as b4
-- left join student on b4.student_id = student.sid;

-- 八、查詢學過「葉平」老師所教的全部課的同窗的學號、姓名；
-- SELECT sname,sid from student where sid in (select student_id from score where course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname='李平老師') group by student_id having count(*) =  (select count(cid) from course left join teacher on course.teacher_id = teacher.tid where tname='李平老師'));

-- 十、查詢有課程成績小於60分的同窗的學號、姓名；
                    # distinct 若是有重複項，只選擇一個
-- select sid,sname from student where sid in (
-- select distinct student_id from score where num<60);

-- 十一、查詢沒有學全全部課的同窗的學號、姓名；
-- select sid,sname from student where sid not in (select student_id from score group by student_id having count(*)=(select count(1) from course));

-- 十二、查詢至少有一門課與學號爲「001」的同窗所學相同的同窗的學號和姓名；
-- select distinct student_id,student.sname from score left join student on score.student_id = student.sid where student_id != 1 and course_id in (select course_id from score where student_id = 1);

--     select student_id,sname, count(course_id)
--         from score left join student on score.student_id = student.sid
--         where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id

-- 1三、查詢至少學過學號爲「001」同窗全部課的其餘同窗學號和姓名；
-- select student_id from score where student_id !=1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(*) >= (select count(course_id) from score where student_id = 1);

-- 1四、查詢和「002」號的同窗學習的課程徹底相同的其餘同窗學號和姓名；
         #1.課程包含 002，且課程數同樣
-- select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
--             select student_id from score  where student_id != 2 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 2)
--         ) and course_id = (select count(1) from score where student_id = 2)


-- 1六、向SC表中插入一些記錄，這些記錄要求符合如下條件：①沒有上過編號「002」課程的同窗學號；②插入「002」號課程的平均成績；
--     思路：
--         因爲insert 支持 
--                 inset into tb1(xx,xx) select x1,x2 from tb2;
--         全部，獲取全部沒上過002課的全部人，獲取002的平均成績
--  
--     insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)
--     from student where sid not in (
--         select student_id from score where course_id = 2)

-- 1七、按平均成績從低到高顯示全部學生的「生物」、「物理」、「體育」三門的課程成績，按以下形式顯示： 學生ID,生物,物理,體育,有效課程數,有效平均分；
-- SELECT
-- student_id,
-- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 1) as 語文,
-- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 2) as 數學,
-- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 3) as 英語
-- from score as s1;

-- 1八、查詢各科成績最高和最低的分：以以下形式顯示：課程ID，最高分，最低分；
-- select course_id,max(num),min(num),cname,case when min(num) < 10 THEN 0 else min(num) end from score left join course on score.course_id = course.cid group by course_id;

-- 1九、按各科平均成績從低到高和及格率的百分數從高到低順序；
-- select course_id,avg(num),sum(case when num<60 then 0 else 1 end),sum(1),sum(case when num<60 then 0 else 1 end)/sum(1) as b from score group by course_id order by avg(num) asc,b desc;

-- 20、課程平均分從高到低顯示（顯示任課老師）；
-- select course_id,cname,tname,avg(if(isnull(score.num),0,score.num)) from score 
-- left join course on score.course_id = course.cid 
-- left join teacher on course.teacher_id = teacher.tid
-- group by course_id order by avg(num) desc;

-- 2一、查詢各科成績前三名的記錄:(不考慮成績並列狀況)
-- select * from
-- (
-- select 
-- student_id,
-- course_id,
-- num,
-- 1,
-- (select num from score as s2 where s2.course_id = s1.course_id group by s2.num order by s2.num desc limit 0,1),
-- (select num from score as s2 where s2.course_id = s1.course_id group by s2.num order by s2.num desc limit 3,1) as cc
-- 
-- from score as s1 
-- ) as B
-- where B.num > B.cc order by B.course_id desc;

-- 2六、查詢同名同姓學生名單，並統計同名人數；
-- select sname,count(1) from student group by sname

-- 2七、查詢每門課程的平均成績，結果按平均成績升序排列，平均成績相同時，按課程號降序排列；
-- select avg(if(isnull(score.num),0,score.num)),course_id from score group by course_id order by avg(num) asc,course_id desc;

-- 2九、查詢課程名稱爲「數學」，且分數低於60的學生姓名和分數；
-- select student.sname,score.num from score 
-- left join course on score.course_id = course.cid
-- left join student on score.student_id = student.sid
-- where course.cname = '生物' and score.num < 60;

-- 3二、查詢選修「李平老師」所授課程的學生中，成績最高的學生姓名及其成績；
-- select student_id,sname,max(num) from score left join student on score.student_id = student.sid where course_id in (
-- select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老師') group by student_id order by max(num) desc limit 0,1;

-- 3四、查詢不一樣課程但成績相同的學生的學號、課程號、學生成績；
-- 此處要了解笛卡兒積
--     select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;
-- 
-- 3八、查詢沒學過「李平老師」老師講授的任一門課程的學生姓名；
    #找出選過李平老師的，而後在學生表 not in 
-- select sid,sname from student where sid not in (select student_id from score where course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where teacher.tname='李平老師') group by student_id)

習題題目及答案

##還能夠選擇查詢語句，不過這個語句結果須要爲一個常量 select age,name,(select count(1) from tb) from tb1; 若是不是常量，須要設定條件，不然會變爲笛卡兒積了。以下所示
# -- 1七、按平均成績從低到高顯示全部學生的「生物」、「物理」、「體育」三門的課程成績，按以下形式顯示： 學生ID,生物,物理,體育,有效課程數,有效平均分；
# -- SELECT
# -- student_id,
# -- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 1) as 語文,
# -- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 2) as 數學,
# -- (select num from score as s2 where s2.student_id=s1.student_id and course_id = 3) as 英語
# -- from score as s1;

'''
select id,(select * from tb1 where tb1.id = 1) from tb2;
此時 子查詢相似一個常量， 每一行就是 id1 常量， id2 常量這樣子
'''
'''
#?????? -- 2一、查詢各科成績前三名的記錄:(不考慮成績並列狀況)
# select * from
# (
# select
# student_id,
# course_id,
# num,
# 1,
# (select num from score as s2 where s2.course_id = s1.course_id group by s2.num order by s2.num desc limit 0,1),
# (select num from score as s2 where s2.course_id = s1.course_id group by s2.num order by s2.num desc limit 3,1) as cc
#
# from score as s1
# ) as B
# where B.num > B.cc order by B.course_id desc;
'''
#重點 s2.student_id=s1.student_id
# SELECT
# student_id,
# (select num from score as s2 where s2.student_id=s1.student_id and course_id = 1) as 語文,
# (select num from score as s2 where s2.student_id=s1.student_id and course_id = 2) as 數學,
# (select num from score as s2 where s2.student_id=s1.student_id and course_id = 3) as 英語
# from score as s1;


# case then 條件 else 字段 END
#select course_id,max(num),min(num),cname,
# case when min(num) < 10 THEN 0 else min(num) end from score left join course on score.course_id = course.cid group by course_id;


#-- 1九、按各科平均成績從低到高和及格率的百分數從高到低順序；
# select course_id,avg(num),sum(case when num<60 then 0 else 1 end),sum(1),sum(case when num<60 then 0 else 1 end)/sum(1) from score group by course_id;


#-- 20、課程平均分從高到低顯示（顯示任課老師）；
# select course_id,cname,tname,avg(if(isnull(score.num),0,score.num)) from score
# left join course on score.course_id = course.cid
# left join teacher on course.teacher_id = teacher.tid
# group by course_id order by avg(num) desc;

#if(isnull(score.num),0,score.num) 若是 score.num是空，那麼就是0，不然是它自己

# -- 3四、查詢不一樣課程但成績相同的學生的學號、課程號、學生成績；
# -- 此處要了解笛卡兒積，鏈接兩張表不加 on 條件，會進行笛卡兒積，就是一張表的首行遍歷鏈接另外一張表全部行，而後第二行繼續遍歷鏈接.....
# --     select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;

答題中幾道不會的題目