繼續sql面試經典50題練習筆記一的練習…
沒有看過sql面試經典50題練習筆記一的朋友建議可以點擊下面的練習先從一開始看起，數據庫測試表和題目來源都有在那篇博客說明~
sql面試經典50題練習筆記一：傳送門 https://blog.csdn.net/maggrect/article/details/102461646
讓我們繼續吧…
16、檢索"01"課程分數小於60，按分數降序排列的學生信息
分析：這道題比較常規，可以先查出01課程分數小於60分的學生id，再根據這個學生id就可以拼湊出想要的結果，
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 ；
我和答案的區別：這道題感覺答案的解法要比我的要好很多，我使用了左連接又用了IN，答案是直接操作兩張表，學習了~~
-- 先查出'01'課程分數小於60的學生的id，這裏我把分數也查了出來，及時爲了看一下而已
SELECT sc.s_id,sc.s_score FROM score sc WHERE sc.s_score<60 AND sc.c_id ='01'
-- 再拼接學生信息
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
-- 參考答案：
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
這道題我做錯了，一拿到這道題我就先去看了課程表，然後就將每個人的課程都限定死爲三門課了，於是就有了以下答案
假如每個人的課程都是固定的，那麼以下解法就是正確的，沒有成績的分數單做0來處理，每個人的課程數是固定的
函數：IFNULL(E,D),E的值如果是null的話就用D的值去替代他
ROUND(E,D),表達式E保留小數點後面D位數字
-- 我的答案
SELECT s.s_id,s.s_name,
IFNULL(b.s_score,0) AS 語文,
IFNULL(c.s_score,0) AS 數學,
IFNULL(d.s_score,0) AS 英語,
ROUND(IFNULL(((IFNULL(b.s_score,0)+IFNULL(c.s_score,0)+IFNULL(d.s_score,0))/3),0),2) AS 平均成績
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 平均成績 DESC
得到的結果：
讓我們來看一下答案的解法：答案的解法是相對比較合理的，沒有該課程成績的人就相當於沒有選該課程，對應的平均成績底數也是不同的！
-- 答案解法
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 語文,
(select s_score from score where s_id=a.s_id and c_id='02') as 數學,
(select s_score from score where s_id=a.s_id and c_id='03') as 英語,
round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC;
結果：
寫了這麼多sql語句，感覺現在大部分場景的sql語句能設計出來了，就是一點一點調出自己想要的效果，要設計出效率更高更準確的sql語句來，還需要對數據庫有更深入的瞭解，接下來的題目我打算暫時不做了，想先繼續學習數據庫優化方面的知識，我會將題目和答案貼在下面，供大家學習。
當然我有空的時候也會來繼續做的，將自己的答案和參考答案做對比並寫出自己的分析過程~
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
–及格爲>=60，中等爲：70-80，優良爲：80-90，優秀爲：>=90
-- 參考答案
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 優良率,
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 優秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
19、按各科成績進行排序，並顯示排名(實現不完全)
– mysql沒有rank函數
- 這道題答案也沒有實現完全
-- 參考答案
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、查詢學生的總成績並進行排名
我的理解：做這道題之前首先需要知道 @i:=@i+1 是什麼意思：
@i:=@i+1的意思可以理解爲，定義了一個變量，然後每次 疊加一，在這裏的作用是生成排名序列號
from後面的 @k:=0,@i:=0,@score:=0 這些爲初始化變量
其次需要理解case when 條件A then 結果一 else 結果二 end 的作用
這裏的意思是加假如條件A成立，那麼就是結果一，否則就是結果二
在這道題中：這裏的作用是用來查看與上一個人分數是否相同，如果相同的話，就使用的上一次的排名（假如學生A和學生B的成績都是100分，那麼兩個人的排名都是第一，不過由於i的值是一直在增加的，所以會沒有了第二名，下一個排名直接是第三名）
參考博客：https://www.cnblogs.com/bjwylpx/p/5345162.html
-- 參考答案
select a.s_id,
//這個i每次增加1
@i:=@i+1 as i,
//當兩次成績不一致的時候，纔會將i賦值給排名(i照常自增)，否則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、查詢不同老師所教不同課程平均分從高到低顯示
--參考答案
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
-- 參考答案
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、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],[0-60]及所佔百分比
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、查詢學生平均成績及其名次
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、查詢各科成績前三名的記錄
– 1.選出b表比a表成績大的所有組
– 2.選出比當前id成績大的 小於三個的
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、查詢每門課程被選修的學生數
select c_id,count(s_id) from score a 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
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、查詢男生、女生人數
select s_sex,COUNT(s_sex) as 人數 from student GROUP BY s_sex
29、查詢名字中含有"風"字的學生信息
select * from student where s_name like '%風%';
30、查詢同名同性學生名單，並統計同名人數
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、查詢1990年出生的學生名單
select s_name from student where s_birth like '1990%'
32、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 ASC
33、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(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、查詢課程名稱爲"數學"，且分數低於60的學生姓名和分數
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 ='數學') and b.s_score<60
35、查詢所有學生的課程及分數情況；
select a.s_id,a.s_name,
SUM(case c.c_name when '語文' then b.s_score else 0 end) as '語文',
SUM(case c.c_name when '數學' then b.s_score else 0 end) as '數學',
SUM(case c.c_name when '英語' then b.s_score else 0 end) as '英語',
SUM(b.s_score) as '總分'
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、查詢任何一門課程成績在70分以上的姓名、課程名稱和分數
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、查詢不及格的課程
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、查詢課程編號爲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 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、求每門課程的學生人數
select count(*) from score GROUP BY c_id;
40、查詢選修"張三"老師所授課程的學生中，成績最高的學生信息及其成績
-- 查詢老師id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='張三'
-- 查詢最高分（可能有相同分數）
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')
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
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、查詢每門功成績最好的前兩名
-- 牛逼的寫法
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、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(*) as total from score
GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
44、檢索至少選修兩門課程的學生學號
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
45、查詢選修了全部課程的學生信息
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
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、查詢本週過生日的學生
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、查詢下週過生日的學生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
49、查詢本月過生日的學生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
50、查詢下月過生日的學生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)