mysql50道题 查询和,Mysql Sql 语句练习题 (50道)
-- 1、查詢'01'課程比'02'課程成績高的學生的信息及課程分數 select st.*,sc.s_score as '語文' ,sc2.s_score '數學' from student stleft join score sc on sc.s_id=st.s_id and sc.c_id='01' left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score>sc2.s_score-- 2、查詢'01'課程比'02'課程成績低的學生的信息及課程分數select st.*,sc.s_score '語文',sc2.s_score '數學' from student stleft join score sc on sc.s_id=st.s_id and sc.c_id='01'left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02'where sc.s_score=60-- 4、查詢平均成績小于60分的同學的學生編號和學生姓名和平均成績 -- (包括有成績的和無成績的)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)) end ) cjScore from student stleft join score sc on sc.s_id=st.s_idgroup by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL-- 5、查詢所有同學的學生編號、學生姓名、選課總數、所有課程的總成績select st.s_id,st.s_name,count(c.c_id),( case when SUM(sc.s_score) is null or sum(sc.s_score)='' then 0 else SUM(sc.s_score) end) from student stleft join score sc on sc.s_id =st.s_id left join course c on c.c_id=sc.c_idgroup by st.s_id-- 6、查詢'李'姓老師的數量 select t.t_name,count(t.t_id) from teacher tgroup by t.t_id having t.t_name like '李%'; -- 7、查詢學過'張三'老師授課的同學的信息 select st.* from student st left join score sc on sc.s_id=st.s_idleft join course c on c.c_id=sc.c_idleft join teacher t on t.t_id=c.t_id where t.t_name='張三'-- 8、查詢沒學過'張三'老師授課的同學的信息 -- 張三老師教的課 select c.* from course c left join teacher t on t.t_id=c.t_id where t.t_name='張三' -- 有張三老師課成績的st.s_id select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.t_name='張三') -- 不在上面查到的st.s_id的學生信息,即沒學過張三老師授課的同學信息 select st.* 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 t.t_id=c.t_id where t.t_name='張三') )-- 9、查詢學過編號為'01'并且也學過編號為'02'的課程的同學的信息select st.* from student st inner join score sc on sc.s_id= st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='01'where st.s_id in (select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_idinner join course c2 on c2.c_id=sc2.c_id and c2.c_id='02')網友提供的思路(厲害呦~):SELECT st.*FROM student stINNER JOIN score sc ON sc.`s_id`=st.`s_id`GROUP BY st.`s_id`HAVING SUM(IF(sc.`c_id`='01' OR sc.`c_id`='02' ,1,0))>1-- 10、查詢學過編號為'01'但是沒有學過編號為'02'的課程的同學的信息select st.* from student st inner join score sc on sc.s_id= st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='01'where st.s_id not in (select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_idinner join course c2 on c2.c_id=sc2.c_id and c2.c_id='02')-- 11、查詢沒有學全所有課程的同學的信息 -- 太復雜,下次換一種思路,看有沒有簡單點方法 -- 此處思路為查學全所有課程的學生id,再內聯取反面select * from student where s_id not in (select st.s_id from student st inner join score sc on sc.s_id = st.s_id and sc.c_id='01'where st.s_id in (select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id='02') and st.s_id in (select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id='03'))-- 來自一樓網友的思路,左連接,根據學生id分組過濾掉 數量小于 課程表中總課程數量的結果(show me his code),簡潔不少。select st.* from Student stleft join Score Son st.s_id= S.s_idgroup by st.s_idhaving count(c_id)=2)group by st.s_id-- 16、檢索'01'課程分數小于60,按分數降序排列的學生信息select st.*,sc.s_score from student st inner join score sc on sc.s_id=st.s_id and sc.c_id='01' and sc.s_score<60order by sc.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 -- 可加round,case when then else end 使顯示更完美select st.s_id,st.s_name,avg(sc4.s_score) '平均分',sc.s_score '語文',sc2.s_score '數學',sc3.s_score '英語' from student stleft join score sc on sc.s_id=st.s_id and sc.c_id='01'left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02'left join score sc3 on sc3.s_id=st.s_id and sc3.c_id='03'left join score sc4 on sc4.s_id=st.s_idgroup by st.s_id order by SUM(sc4.s_score) 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:>=90select c.c_id,c.c_name,max(sc.s_score) '最高分',MIN(sc2.s_score) '最低分',avg(sc3.s_score) '平均分' ,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) '及格率',((select count(s_id) from score where s_score>=70 and s_score<80 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) '中等率',((select count(s_id) from score where s_score>=80 and s_score<90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) '優良率',((select count(s_id) from score where s_score>=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) '優秀率'from course cleft 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-- 19、按各科成績進行排序,并顯示排名(實現不完全)-- mysql沒有rank函數-- 加@score是為了防止用union all 后打亂了順序select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_idwhere c.c_id='01' order by sc.s_score desc) c1 ,(select @i:=0) aunion all select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_idwhere c.c_id='02' order by sc.s_score desc) c2 ,(select @ii:=0) aa union allselect c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_idwhere c.c_id='03' order by sc.s_score desc) c3;set @iii=0;-- 20、查詢學生的總成績并進行排名select st.s_id,st.s_name,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end) from student stleft join score sc on sc.s_id=st.s_idgroup by st.s_id order by sum(sc.s_score) desc-- 21、查詢不同老師所教不同課程平均分從高到低顯示 select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t left join course c on c.t_id=t.t_id left join score sc on sc.c_id =c.c_idgroup by t.t_idorder by avg(sc.s_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select a.* from (select st.*,c.c_id,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id =sc.c_id and c.c_id='01'order by sc.s_score desc LIMIT 1,2 ) aunion allselect b.* from (select st.*,c.c_id,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id =sc.c_id and c.c_id='02'order by sc.s_score desc LIMIT 1,2) bunion allselect c.* from (select st.*,c.c_id,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id =sc.c_id and c.c_id='03'order by sc.s_score desc LIMIT 1,2) c-- 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]及所占百分比select c.c_id,c.c_name ,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100 and sc.s_score>80)/(select count(1) from score sc where sc.c_id=c.c_id )) '100-85',((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85 and sc.s_score>70)/(select count(1) from score sc where sc.c_id=c.c_id )) '85-70',((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70 and sc.s_score>60)/(select count(1) from score sc where sc.c_id=c.c_id )) '70-60',((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60 and sc.s_score>=0)/(select count(1) from score sc where sc.c_id=c.c_id )) '60-0'from course c order by c.c_id-- 24、查詢學生平均成績及其名次 set @i=0;select a.*,@i:=@i+1 from (select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2) '平均分' from student stleft join score sc on sc.s_id=st.s_idgroup by st.s_id order by sc.s_score desc) a-- 25、查詢各科成績前三名的記錄select a.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='01' order by sc.s_score desc LIMIT 0,3) aunion all select b.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='02' order by sc.s_score desc LIMIT 0,3) bunion allselect c.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='03' order by sc.s_score desc LIMIT 0,3) c-- 26、查詢每門課程被選修的學生數 select c.c_id,c.c_name,count(1) from course c left join score sc on sc.c_id=c.c_idinner join student st on st.s_id=c.c_idgroup by st.s_id-- 27、查詢出只有兩門課程的全部學生的學號和姓名select st.s_id,st.s_name from student st left join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id group by st.s_id having count(1)=2-- 28、查詢男生、女生人數select st.s_sex,count(1) from student st group by st.s_sex-- 29、查詢名字中含有'風'字的學生信息select st.* from student st where st.s_name like '%風%';-- 30、查詢同名同性學生名單,并統計同名人數 select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1-- 31、查詢1990年出生的學生名單select st.* from student st where st.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.c_id,c.c_name,avg(sc.s_score) from course cinner join score sc on sc.c_id=c.c_id group by c.c_id order by avg(sc.s_score) desc,c.c_id asc-- 33、查詢平均成績大于等于85的所有學生的學號、姓名和平均成績select st.s_id,st.s_name,avg(sc.s_score) from student stleft join score sc on sc.s_id=st.s_idgroup by st.s_id having avg(sc.s_score)>=85-- 34、查詢課程名稱為'數學',且分數低于60的學生姓名和分數 select st.s_id,st.s_name,sc.s_score from student stinner join score sc on sc.s_id=st.s_id and sc.s_score<60inner join course c on c.c_id=sc.c_id and c.c_name ='數學' -- 35、查詢所有學生的課程及分數情況;select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idleft join course c on c.c_id =sc.c_idorder by st.s_id,c.c_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 st2.s_id,st2.s_name,c2.c_name,sc2.s_score from student st2left join score sc2 on sc2.s_id=st2.s_idleft join course c2 on c2.c_id=sc2.c_id where st2.s_id in(select st.s_id from student st left join score sc on sc.s_id=st.s_id group by st.s_id having min(sc.s_score)>=70)order by s_id-- 37、查詢不及格的課程select st.s_id,c.c_name,st.s_name,sc.s_score from student stinner join score sc on sc.s_id=st.s_id and sc.s_score<60inner join course c on c.c_id=sc.c_id -- 38、查詢課程編號為01且課程成績在80分以上的學生的學號和姓名select st.s_id,st.s_name,sc.s_score from student stinner join score sc on sc.s_id=st.s_id and sc.c_id='01' and sc.s_score>=80-- 39、求每門課程的學生人數select c.c_id,c.c_name,count(1) from course cinner join score sc on sc.c_id=c.c_idgroup by c.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 select st.*,c.c_name,sc.s_score,t.t_name from student stinner join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id inner join teacher t on t.t_id=c.t_id and t.t_name='張三'order by sc.s_score desclimit 0,1-- 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 select st.s_id,st.s_name,sc.c_id,sc.s_score from student st left join score sc on sc.s_id=st.s_idleft join course c on c.c_id=sc.c_idwhere (select count(1) from student st2 left join score sc2 on sc2.s_id=st2.s_idleft join course c2 on c2.c_id=sc2.c_idwhere sc.s_score=sc2.s_score and c.c_id!=c2.c_id )>1-- 42、查詢每門功成績最好的前兩名 select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='01'order by sc.s_score desc limit 0,2) aunion allselect b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='02'order by sc.s_score desc limit 0,2) bunion allselect c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='03'order by sc.s_score desc limit 0,2) c -- 借鑒(更準確,漂亮): 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 sc.c_id,count(1) from score scleft join course c on c.c_id=sc.c_idgroup by c.c_id having count(1)>5order by count(1) desc,sc.c_id asc-- 44、檢索至少選修兩門課程的學生學號 select st.s_id from student st left join score sc on sc.s_id=st.s_idgroup by st.s_id having count(1)>=2-- 45、查詢選修了全部課程的學生信息select st.* from student st left join score sc on sc.s_id=st.s_idgroup by st.s_id having count(1)=(select count(1) from course)-- 46、查詢各學生的年齡 select st.*,timestampdiff(year,st.s_birth,now()) from student st-- 47、查詢本周過生日的學生 -- 此處可能有問題,week函數取的為當前年的第幾周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期幾(%w), -- 再判斷本周是否會持續到下一個月進行判斷,太麻煩,不會寫select st.* from student st where week(now())=week(date_format(st.s_birth,'%Y%m%d'))-- 48、查詢下周過生日的學生select st.* from student st where week(now())+1=week(date_format(st.s_birth,'%Y%m%d'))-- 49、查詢本月過生日的學生select st.* from student st where month(now())=month(date_format(st.s_birth,'%Y%m%d'))-- 50、查詢下月過生日的學生 -- 注意:當 當前月為12時,用month(now())+1為13而不是1,可用timestampadd()函數或mod取模select st.* from student st where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,'%Y%m%d'))-- 或select st.* from student st where (month(now()) + 1) mod 12= month(date_format(st.s_birth,'%Y%m%d'))
總結
以上是生活随笔為你收集整理的mysql50道题 查询和,Mysql Sql 语句练习题 (50道)的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: 谷歌旗下自动驾驶公司 Waymo 第二轮
- 下一篇: 虚幻引擎 5 打造,开放世界 MMO 游