sql面试经典50题练习笔记二

继续sql面试经典50题练习笔记一的练习…

没有看过sql面试经典50题练习笔记一的朋友建议可以点击下面的练习先从一开始看起，数据库测试表和题目来源都有在那篇博客说明~

sql面试经典50题练习笔记一：传送门 https://blog.csdn.net/maggrect/article/details/102461646

让我们继续吧…

16、检索"01"课程分数小于60，按分数降序排列的学生信息
分析：这道题比较常规，可以先查出01课程分数小于60分的学生id，再根据这个学生id就可以拼凑出想要的结果，
注意：
升序：select 列名 from 表名 order by 表中的字段 asc；
降序：select 列名 from 表名 order by 表中的字段 desc ；
我和答案的区别：这道题感觉答案的解法要比我的要好很多，我使用了左连接又用了IN，答案是直接操作两张表，学习了~~


-- 先查出'01'课程分数小于60的学生的id，这里我把分数也查了出来，及时为了看一下而已
SELECT sc.s_id,sc.s_score FROM score sc WHERE sc.s_score<60 AND sc.c_id ='01'

-- 再拼接学生信息
SELECT s.*, a.s_score FROM student s LEFT JOIN 
score a ON s.s_id = a.s_id WHERE s.s_id IN
(SELECT sc.s_id FROM score sc WHERE sc.s_score<60 AND sc.c_id ='01') 
GROUP BY s.s_id 
ORDER BY a.s_score DESC

-- 参考答案：
select a.*,b.c_id,b.s_score from 
    student a,score b 
where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.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
这道题我做错了，一拿到这道题我就先去看了课程表，然后就将每个人的课程都限定死为三门课了，于是就有了以下答案

假如每个人的课程都是固定的，那么以下解法就是正确的，没有成绩的分数单做0来处理，每个人的课程数是固定的

函数：IFNULL(E,D),E的值如果是null的话就用D的值去替代他
ROUND(E,D),表达式E保留小数点后面D位数字

-- 我的答案
SELECT s.s_id,s.s_name,
IFNULL(b.s_score,0) AS 语文,
IFNULL(c.s_score,0) AS 数学,
IFNULL(d.s_score,0) AS 英语,
ROUND(IFNULL(((IFNULL(b.s_score,0)+IFNULL(c.s_score,0)+IFNULL(d.s_score,0))/3),0),2) AS 平均成绩 
FROM student s
LEFT JOIN score b ON s.s_id = b.s_id AND b.c_id =01
LEFT JOIN score c ON c.s_id = s.s_id AND c.c_id =02
LEFT JOIN score d ON s.s_id = d.s_id AND d.c_id = 03
GROUP BY s.s_id ORDER BY 平均成绩 DESC

得到的结果：

让我们来看一下答案的解法：答案的解法是相对比较合理的，没有该课程成绩的人就相当于没有选该课程，对应的平均成绩底数也是不同的！

-- 答案解法
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
(select s_score from score where s_id=a.s_id and c_id='02') as 数学,
(select s_score from score where s_id=a.s_id and c_id='03') as 英语,
round(avg(s_score),2) as 平均分 from score a  GROUP BY a.s_id ORDER BY 平均分 DESC;

结果：

写了这么多sql语句，感觉现在大部分场景的sql语句能设计出来了，就是一点一点调出自己想要的效果，要设计出效率更高更准确的sql语句来，还需要对数据库有更深入的了解，接下来的题目我打算暂时不做了，想先继续学习数据库优化方面的知识，我会将题目和答案贴在下面，供大家学习。

当然我有空的时候也会来继续做的，将自己的答案和参考答案做对比并写出自己的分析过程~

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

-- 参考答案
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
    ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
    ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
    ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
    ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
    from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

19、按各科成绩进行排序，并显示排名(实现不完全)
– mysql没有rank函数

这道题答案也没有实现完全

-- 参考答案
 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 WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
    union
    select a.s_id,a.c_id,
        @i:=@i +1 as i,
        @k:=(case when @score=a.s_score then @k else @i end) as rank,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
    union
    select a.s_id,a.c_id,
        @i:=@i +1 as i,
        @k:=(case when @score=a.s_score then @k else @i end) as rank,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s

20、查询学生的总成绩并进行排名
我的理解：做这道题之前首先需要知道 @i:=@i+1 是什么意思：
@i:=@i+1的意思可以理解为，定义了一个变量，然后每次叠加一，在这里的作用是生成排名序列号
from后面的 @k:=0,@i:=0,@score:=0 这些为初始化变量

其次需要理解case when 条件A then 结果一 else 结果二 end 的作用
这里的意思是加假如条件A成立，那么就是结果一，否则就是结果二

在这道题中：这里的作用是用来查看与上一个人分数是否相同，如果相同的话，就使用的上一次的排名（假如学生A和学生B的成绩都是100分，那么两个人的排名都是第一，不过由于i的值是一直在增加的，所以会没有了第二名，下一个排名直接是第三名）
参考博客：https://www.cnblogs.com/bjwylpx/p/5345162.html

-- 参考答案
select a.s_id,
	//这个i每次增加1
    @i:=@i+1 as i,
    //当两次成绩不一致的时候，才会将i赋值给排名(i照常自增)，否则使用的上一次的排名
    @k:=(case when @score=a.sum_score then @k else @i end) as rank,
    @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

21、查询不同老师所教不同课程平均分从高到低显示

--参考答案
select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
        left join score b on a.c_id=b.c_id 
        left join teacher c on a.t_id=c.t_id
        GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_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

-- 参考答案
 select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3;

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 distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
	left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
        ROUND(100*(SUM(case when s_score >85 and s_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 s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
        ROUND(100*(SUM(case when s_score >70 and s_score <=85 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 s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
        ROUND(100*(SUM(case when s_score >60 and s_score <=70 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 (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
        ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
         from score GROUP BY c_id)e on a.c_id=e.c_id
     left join course f on a.c_id = f.c_id

24、查询学生平均成绩及其名次

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(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;

25、查询各科成绩前三名的记录
– 1.选出b表比a表成绩大的所有组
– 2.选出比当前id成绩大的小于三个的

 select a.s_id,a.c_id,a.s_score from score a 
       left join score b on a.c_id = b.c_id and a.s_score<b.s_score
       group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
       ORDER BY a.c_id,a.s_score DESC

26、查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

27、查询出只有两门课程的全部学生的学号和姓名

 select s_id,s_name from student where s_id in(
     select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

28、查询男生、女生人数

select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex

29、查询名字中含有"风"字的学生信息

 select * from student where s_name like '%风%';

30、查询同名同性学生名单，并统计同名人数

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

31、查询1990年出生的学生名单

select s_name from student where 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_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,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 a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
 left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_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 a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id 
where b.c_id=(select c_id from course where c_name ='数学') and b.s_score<60

35、查询所有学生的课程及分数情况；

 select a.s_id,a.s_name,
       SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
       SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
       SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
       SUM(b.s_score) as  '总分'
        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

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 a.s_name,b.c_name,c.s_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.s_score>=70

37、查询不及格的课程

select a.s_id,a.c_id,b.c_name,a.s_score from score a 
left join course b on a.c_id = b.c_id where a.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 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.s_score>80

39、求每门课程的学生人数

select count(*) from score GROUP BY 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

-- 查询老师id   
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查询最高分（可能有相同分数）
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='张三')
 and b.s_score in (select MAX(s_score) from score where c_id='02')

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 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

42、查询每门功成绩最好的前两名

-- 牛逼的写法
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

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_id,count(*) as total from score 
GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

44、检索至少选修两门课程的学生学号

select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

45、查询选修了全部课程的学生信息

 select * from student where s_id in(        
select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))

46、查询各学生的年龄

-- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
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;

47、查询本周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
 select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

48、查询下周过生日的学生

 select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

49、查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

50、查询下月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

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