MySql 语句练习50题

-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
SELECT st.*,sc1.s_score "语文",sc2.s_score "数学" FROM student st
left join score sc1 on st.s_id = sc1.s_id and sc1.c_id ="01"
left join score sc2 on st.s_id = sc2.s_id and sc2.c_id ="02"
where sc1.s_score > sc2.s_score


-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
SELECT st.*,sc.s_score "语文", sc1.s_score "数学" from student st
left join score sc on st.s_id = sc.s_id and sc.c_id="01"
left join score sc1 on st.s_id =sc1.s_id and sc1.c_id="02"
where sc.s_score < sc1.s_score


-- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select st.s_id,st.s_name,Round(avg(sc.s_score),2) "平均分" from student st 
left join score sc on  st.s_id = sc.s_id 
group by st.s_id
having avg(sc.s_score) > 60

-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
        -- (包括有成绩的和无成绩的)
select st.s_id,st.s_name,(case WHEN Round(avg(sc.s_score),2) is null then 0 else Round(avg(sc.s_score),2)  end) "平均分"  from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id 
HAVING avg(sc.s_score) < 60 or avg(sc.s_score)  is null
                
-- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
select st.s_id,st.s_name,count(c.c_id) "课程总数", 
(case WHEN sum(sc.s_score) is null or sum(sc.s_score) = "" then 0 else sum(sc.s_score) end) "总成绩" 
 from student st 
left join score sc on st.s_id = sc.s_id
left join course c on sc.c_id = c.c_id
group by st.s_id


-- 6、查询"李"姓老师的数量
select t.t_name,count(1) from teacher t
group by t.t_name
having t.t_name like "李%"



-- 7、查询学过"张三"老师授课的同学的信息
select st.* from student st
left join score sc on st.s_id = sc.s_id
left join course c on sc.c_id = c.c_id
left join teacher t on c.t_id = t.t_id
where t.t_name = "张三"


-- 8、查询没学过"张三"老师授课的同学的信息 
 -- 张三老师教的课
select s.* from student s  
where s.s_id not in (
    select st.s_id from student st
    left join score sc on st.s_id = sc.s_id
    left join course c on sc.c_id = c.c_id
    left join teacher t on c.t_id = t.t_id
    where t.t_name = "张三"
)

-- 1.先查询张三老师教的所有课程
-- 2.然后通过所有可能,拿到成绩表对应的学生id
-- 3.去掉所有张三老师教过的学生id

-- 第1步
select * from course c 
left join teacher t on c.t_id = t.t_id
where t.t_name = "张三"
-- 第2步
select * from score sc  
where sc.c_id in (
    select c.c_id from course c 
    left join teacher t on c.t_id = t.t_id
    where t.t_name = "张三"
)
-- 第3步
select * from student st
where st.s_id not in (
    select sc.s_id from score sc  
    where sc.c_id in (
        select c.c_id from course c 
        left join teacher t on c.t_id = t.t_id
        where t.t_name = "张三"
    )
)


-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
-- 先查询出"02"编号的学生,然后再重新查询学生表,将查询出来的学生当作条件,再查询"01"编号
select st.* from student st
LEFT join score sc on st.s_id = sc.s_id
left join course c on sc.c_id = c.c_id
WHERE c.t_id = "01" and st.s_id in (
    select s.s_id from student s
    LEFT join score sc on s.s_id = sc.s_id
    left join course c on sc.c_id = c.c_id
    WHERE c.t_id = "02"
)

-- 查询成绩表,使用if条件,判断课程id = 01 或者 =02 就 +1
-- 使用聚合函数必须分组 / 多行数据合并,必须分组
select st.* from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id
having sum(if(sc.c_id = "01" or sc.c_id = "02",1,0)) > 1



-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
 select st.* from student st
 left join score sc on st.s_id = sc.s_id
 where sc.c_id = "01" and st.s_id not in(
         select st.s_id from student st
         left join score sc on st.s_id = sc.s_id
         where sc.c_id = "02"
 )



-- 11、查询没有学全所有课程的同学的信息
 -- 太复杂,下次换一种思路,看有没有简单点方法
 -- 此处思路为查学全所有课程的学生id,再内联取反面
 -- 方式一:
select s.s_id from student s
where s.s_id not in (
    select st.s_id from student st
    left join score sc on st.s_id = sc.s_id
    left join score sc1 on st.s_id = sc1.s_id
    left join score sc2 on st.s_id = sc2.s_id
    where sc.c_id ="01" and  sc1.c_id ="03" and  sc2.c_id ="02"
    group by st.s_id
)
 -- 方式二:
select st.* from student st
where st.s_id not in(
    select s.s_id from student s
    left join score sc on s.s_id = sc.s_id 
    where sc.c_id = "01" and s.s_id in(
        select s2.s_id from student s2
        left join score sc on s2.s_id = sc.s_id 
        where sc.c_id = "02" and s2.s_id in(
            select s3.s_id from student s3
            left join score sc on s3.s_id = sc.s_id 
            where sc.c_id = "03"
        )
    )
)

-- 方式三:更灵活  先回课程的总数量,然后,成绩表通过学生id分组,再获取学生的课程数量,进行于课程表的总数量做对比
select st.* from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id
having count(c_id) < (select count(1) from course)



-- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息\
select s.* from student s
left join score scr on s.s_id = scr.s_id
where scr.c_id in (
    select sc.c_id from student st
    left join score sc on st.s_id = sc.s_id
    where st.s_id ="01"
)
group by s.s_id

                
-- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息        
select s.* from student s
left join score s1 on s.s_id = s1.s_id
group by s.s_id 
having GROUP_CONCAT(s1.c_id) = 
(
select GROUP_CONCAT(sc.c_id) from student st 
left join score sc on st.s_id = sc.s_id
WHERE st.s_id = "01")


-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select st.s_name from student st
left join score sc  on st.s_id = sc.s_id
where st.s_id not in(
    select s1.s_id from score  s1
    left join course c on s1.c_id = c.c_id
    left join teacher t on c.t_id = t.t_id
    where t.t_name = "张三"
    group by st.s_id
)
group by st.s_id


-- 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
select st.s_id, st.s_name,avg(sc.s_score) from student st
left join score sc on st.s_id = sc.s_id 
where st.s_id  in (
    select s1.s_id from score s1
    where s1.s_score < 60 or s1.s_score is null
    group by s1.s_id
    having count(s1.s_id) >= 2
)
group by st.s_id


-- 16、检索"01"课程分数小于60,按分数降序排列的学生信息
select st.*,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
where sc.c_id = "01" and sc.s_score < 60
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
 -- 可加round,case when then else end 使显示更完美
 select st.*,sc.s_score "数学", sc2.s_score "语文", sc3.s_score "英语", avg(sc4.s_score) "平均分" from student st 
 left join score sc on st.s_id = sc.s_id and sc.c_id="01"
 left join score sc2 on st.s_id = sc2.s_id and sc2.c_id="02"
 left join score sc3 on st.s_id = sc3.s_id and sc3.c_id="03"
 left join score sc4 on st.s_id = sc4.s_id   -- 不带查询条件,当前表所有数据
 GROUP BY st.s_id
 order by avg(sc4.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
select c.c_id "课程ID", c.c_name "课程名称", max(sc.s_score) "最高分", min(sc2.s_score) "最低分", avg(sc3.s_score) "平均分"
, ((SELECT count(s_score) from score where s_score >= 60 and c_id = c.c_id) / (SELECT count(s_score) from score  where c_id = c.c_id)) "及格率"
, ((SELECT count(s_score) from score where s_score >= 70 and s_score < 80 and c_id = c.c_id) / (SELECT count(s_score) from score  where c_id = c.c_id))  "中等率"
, ((SELECT count(s_score) from score where s_score >= 80 and s_score < 90 and c_id = c.c_id) / (SELECT count(s_score) from score  where c_id = c.c_id)) "优良率"
, ((SELECT count(s_score) from score where s_score >= 90 and c_id = c.c_id) / (SELECT count(s_score) from score  where c_id = c.c_id)) "优秀率"
from course c 
left join score sc on sc.c_id = c.c_id
left join score sc2 on sc2.c_id = c.c_id
left join score sc3 on sc3.c_id = c.c_id
group by c.c_id


-- 20、查询学生的总成绩并进行排名
select st.*,(case WHEN sum(sc.s_score) is null then 0  else sum(sc.s_score) end) "总成绩" from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id
order by sum(sc.s_score) desc


-- 21、查询不同老师所教不同课程平均分从高到低显示 
select t.t_id,t.t_name,c.c_name, avg(sc.s_score) from score sc
left join course c on sc.c_id = c.c_id
left join teacher t on c.t_id = t.t_id
group by t.t_id,c.c_name
order by avg(sc.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
select s.* from (
select a.* from
(select st.*,c.c_id, c.c_name,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
inner join course c on sc.c_id = c.c_id and c.c_id = "01"
order by sc.s_score desc
limit 1,2) a
UNION ALL
select b.* from
(select st.*,c.c_id, c.c_name,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
inner join course c on sc.c_id = c.c_id and c.c_id = "02"
order by sc.s_score desc
limit 1,2) b
UNION ALL
select c.* from 
(select st.*,c.c_id, c.c_name,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
inner join course c on sc.c_id = c.c_id and c.c_id = "03"
order by sc.s_score desc
limit 1,2) c
) s


-- 23、统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
select c.c_id,c.c_name,
((select count(1) from score s1 where s1.s_score BETWEEN 85 and 100 and s1.c_id = c.c_id) / (select count(1) from score s1 where s1.c_id = c.c_id)) "[100-85]",
((select count(1) from score s1 where s1.s_score BETWEEN 70 and 85 and s1.c_id = c.c_id) / (select count(1) from score s1 where s1.c_id = c.c_id)) "[85-70]",
((select count(1) from score s1 where s1.s_score BETWEEN 60 and 70 and s1.c_id = c.c_id) / (select count(1) from score s1 where s1.c_id = c.c_id)) "[70-60]",
((select count(1) from score s1 where s1.s_score BETWEEN 0 and 60 and s1.c_id = c.c_id) / (select count(1) from score s1 where s1.c_id = c.c_id)) "[0-60]"
from course c GROUP BY c.c_id


-- 24、查询学生平均成绩及其名次
select st.*,sum(sc.s_score) "总成绩" from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id
order by sum(sc.s_score) desc

-- 25、查询各科成绩前三名的 学生信息及该课程成绩
select a.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id 
inner join course c on sc.c_id = c.c_id and c.c_id ="01"
order by sc.s_score desc
limit 0,3) a
union all
select b.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id 
inner join course c on sc.c_id = c.c_id and c.c_id ="02"
order by sc.s_score desc
limit 0,3 ) b
union all
select c.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id 
inner join course c on sc.c_id = c.c_id and c.c_id ="03"
order by sc.s_score desc
limit 0,3) c


-- 26、查询每门课程被选修的学生数
select c.c_id,c.c_name,count(sc.c_id) from score sc 
left join course c on sc.c_id = c.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
select st.*,count(sc.c_id) from student st
left join score sc on st.s_id = sc.s_id 
group by st.s_id 
having count(sc.c_id) = 2


-- 28、查询男生、女生人数
select st.s_sex,count(st.s_id) from student st
GROUP BY st.s_sex


-- 29、查询名字中含有"风"字的学生信息
select st.* from student st 
where st.s_name like "%%"


-- 30、查询同名同性学生名单,并统计同名人数 
select st.*,count(1) from student st
group by st.s_name,st.s_sex
having count(1) > 1


-- 31、查询1990年出生的学生名单
select st.* from student st
where st.s_birth like "1990%"


-- 32、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列 
select c.c_name, avg(sc.s_score) from score sc 
left join course c on sc.c_id = c.c_id
group by c.c_name
order by avg(sc.s_score) desc, c.c_id asc

-- 33、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
select st.*,avg(sc.s_score) from student st
left join score sc on st.s_id = sc.s_id 
GROUP BY st.s_id
having avg(sc.s_score) > 85


-- 34、查询课程名称为"数学",且分数低于60的学生姓名和分数 
select st.*,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
left join course c on sc.c_id = c.c_id
where sc.s_score < 60 and c.c_name = "数学"


-- 35、查询所有学生的课程及分数情况;
select st.*,c.c_name,sc.s_score from student st
left join score sc on st.s_id = sc.s_id 
left join course c on c.c_id = sc.c_id
order by st.s_id,c.c_name



-- 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
select st.*,c.c_name,sc.s_score from student st
left join score sc on st.s_id = sc.s_id
left join course c on sc.c_id = c.c_id
where sc.s_id in (
select sc2.s_id from  score sc2 
left join course c2 on sc2.c_id = c2.c_id
where sc.s_score >= 70 
group by sc2.s_id
)


 -- 37、查询不及格的课程
 select st.s_id,st.s_name,sc.s_score,c.c_name from student st 
 left join score sc on st.s_id = sc.s_id 
 left join course c on sc.c_id = c.c_id
 where sc.s_score < 60
 
 
 -- 38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
 select st.*,c.c_name,sc.s_score from student st 
 left join score sc on st.s_id = sc.s_id
 left join course c on sc.c_id = c.c_id 
 where sc.c_id = "01" and sc.s_score >= 80
 
 
 -- 39、求每门课程的学生人数
 select c.c_name,count(1) from score sc
 left join course c on sc.c_id = c.c_id
 group by c.c_id
 
 
-- 40、查询选修"张三"老师所授课程的学生中,成绩最高的学生信息及其成绩
select st.*,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id 
left join course c on sc.c_id = c.c_id 
left join teacher t on c.t_id = t.t_id
where t.t_name = "张三" 
order by sc.s_score desc
limit 0,1


-- 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
select st.*,c.c_id,c.c_name from student st 
left join score sc on st.s_id = sc.s_id 
left join course c on sc.c_id = c.c_id
where st.s_id in (
    select st2.s_id from student st2 
    left join score sc2 on st2.s_id = sc2.s_id 
    where sc.s_score = sc2.s_score and sc.c_id != sc2.c_id
)

-- 42、查询每门功成绩最好的前两名
select * from (
select st.*,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id
left join course c on c.c_id = sc.c_id
where c.c_id = "01"
order by sc.s_score desc
limit 0,2) a
union all
select * from (
select st.*,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id
left join course c on c.c_id = sc.c_id 
where c.c_id = "02"
order by sc.s_score desc
limit 0,2) b
union all
select * from (
select st.*,c.c_name,sc.s_score from student st 
left join score sc on st.s_id = sc.s_id
left join course c on c.c_id = sc.c_id 
where c.c_id = "03"
order by sc.s_score desc
limit 0,2) c


-- 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
select c.c_id,c.c_name,count(sc.c_id) from score sc 
left join course c on sc.c_id = c.c_id 
group by c.c_id 
having count(sc.c_id) > 5
order by count(sc.c_id) desc,c.c_id asc




-- 44、检索至少选修两门课程的学生学号 
select st.*,count(sc.c_id) from student st 
left join score sc on st.s_id = sc.s_id 
group by st.s_id
having count(sc.c_id) >= 2


-- 45、查询选修了全部课程的学生信息
select st.*,count(sc.c_id) from student st 
left join score sc on st.s_id = sc.s_id
group by st.s_id
having count(sc.c_id) = (select count(1) from course)


-- 46、查询各学生的年龄
select st.s_name,timestampdiff(year,st.s_birth,now()) from student st         --

-- 47、查询本周过生日的学生
  -- 此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w),
  -- 再判断本周是否会持续到下一个月进行判断,太麻烦,不会写
    
select st.*,week(st.s_birth) from student st
where week(now()) = week(DATE_FORMAT(s_birth,"%y%m%d"))

-- 48、查询下周过生日的学生
select st.* from student st 
where week(now())+1=week(date_format(st.s_birth,'%Y%m%d')) 


-- 49、查询本月过生日的学生
select st.*, month(now()) from student st
where month(now()) = month(date_format(st.s_birth,"%y%m%d"))


-- 50、查询下月过生日的学生
 -- 注意:当 当前月为12时,用month(now())+1为13而不是1,可用timestampadd()函数或mod取模
select st.*, month(now()) from student st
where if(month(now()) +1 = 13,1,month(now()) +1) = month(date_format(st.s_birth,"%y%m%d"))

 

 

 

 

转载至: https://www.cnblogs.com/kangxinxin/p/11585935.html

posted @ 2020-05-15 22:35  陌笠人灬苼  阅读(525)  评论(0编辑  收藏  举报