為了練習sql語句,在網上找了一些題,自己做了一遍,收益頗多.很多地方換一種思路,有更好的寫法,歡迎指正.
題目地址:https://blog.csdn.net/fashion2014/article/details/78826299 ,他有更好的寫法
表名和字段
–1.學生表
Student(s_id,s_name,s_birth,s_sex) –學生編號,學生姓名, 出生年月,學生性別
–2.課程表
Course(c_id,c_name,t_id) – –課程編號, 課程名稱, 教師編號
–3.教師表
Teacher(t_id,t_name) –教師編號,教師姓名
–4.成績表
Score(s_id,c_id,s_score) –學生編號,課程編號,分數
測試數據
--建表 --學生表 CREATE TABLE `Student`( `s_id` VARCHAR(20), `s_name` VARCHAR(20) NOT NULL DEFAULT '', `s_birth` VARCHAR(20) NOT NULL DEFAULT '', `s_sex` VARCHAR(10) NOT NULL DEFAULT '', PRIMARY KEY(`s_id`) ); --課程表 CREATE TABLE `Course`( `c_id` VARCHAR(20), `c_name` VARCHAR(20) NOT NULL DEFAULT '', `t_id` VARCHAR(20) NOT NULL, PRIMARY KEY(`c_id`) ); --教師表 CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(`t_id`) ); --成績表 CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`) ); --插入學生表測試數據 insert into Student values('01' , '趙雷' , '1990-01-01' , '男'); insert into Student values('02' , '錢電' , '1990-12-21' , '男'); insert into Student values('03' , '孫風' , '1990-05-20' , '男'); insert into Student values('04' , '李雲' , '1990-08-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '女'); insert into Student values('06' , '吳蘭' , '1992-03-01' , '女'); insert into Student values('07' , '鄭竹' , '1989-07-01' , '女'); insert into Student values('08' , '王菊' , '1990-01-20' , '女'); --課程表測試數據 insert into Course values('01' , '語文' , '02'); insert into Course values('02' , '數學' , '01'); insert into Course values('03' , '英語' , '03'); --教師表測試數據 insert into Teacher values('01' , '張三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五'); --成績表測試數據 insert into Score values('01' , '01' , 80); insert into Score values('01' , '02' , 90); insert into Score values('01' , '03' , 99); insert into Score values('02' , '01' , 70); insert into Score values('02' , '02' , 60); insert into Score values('02' , '03' , 80); insert into Score values('03' , '01' , 80); insert into Score values('03' , '02' , 80); insert into Score values('03' , '03' , 80); insert into Score values('04' , '01' , 50); insert into Score values('04' , '02' , 30); insert into Score values('04' , '03' , 20); insert into Score values('05' , '01' , 76); insert into Score values('05' , '02' , 87); insert into Score values('06' , '01' , 31); insert into Score values('06' , '03' , 34); insert into Score values('07' , '02' , 89); insert into Score values('07' , '03' , 98);
表數據如下:
student 學生表:
score 分數表:
course課程表:
teacher老師表:
-- 准備條件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否則此種情況下會報錯: -- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo. -- 原因： -- MySQL 5.7.5和up實現了對功能依賴的檢測。如果啟用了only_full_group_by SQL模式(在默認情況下是這樣)， -- 那么MySQL就會拒絕選擇列表、條件或順序列表引用的查詢，這些查詢將引用組中未命名的非聚合列，而不是在功能上依賴於它們。 -- (在5.7.5之前，MySQL沒有檢測到功能依賴項，only_full_group_by在默認情況下是不啟用的。關於前5.7.5行為的描述，請參閱MySQL 5.6參考手冊。) -- 執行以下個命令，可以查看 sql_mode 的內容。 SHOW SESSION VARIABLES; SHOW GLOBAL VARIABLES; select @@sql_mode; -- 更改 set global sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION'; set session sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
練習題和sql語句
-- 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 st left 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 st left 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 -- 3、查詢平均成績大於等於60分的同學的學生編號和學生姓名和平均成績 select st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) cjScore from student st left join score sc on sc.s_id=st.s_id group by st.s_id having AVG(sc.s_score)>=60 -- 4、查詢平均成績小於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 st left join score sc on sc.s_id=st.s_id group 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 st left join score sc on sc.s_id =st.s_id left join course c on c.c_id=sc.c_id group by st.s_id -- 6、查詢"李"姓老師的數量 select t.t_name,count(t.t_id) from teacher t group 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_id left join course c on c.c_id=sc.c_id left 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"並且也學過編號為"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_id inner 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_id inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02" ) -- 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_id inner 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_id inner 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" )) -- 12、查詢至少有一門課與學號為"01"的同學所學相同的同學的信息 select distinct st.* from student st left join score sc on sc.s_id=st.s_id where sc.c_id in ( select sc2.c_id from student st2 left join score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' ) -- 13、查詢和"01"號的同學學習的課程完全相同的其他同學的信息 select st.* from student st left join score sc on sc.s_id=st.s_id group by st.s_id having group_concat(sc.c_id) = ( select group_concat(sc2.c_id) from student st2 left join score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' ) -- 14、查詢沒學過"張三"老師講授的任一門課程的學生姓名 select st.s_name from student st where st.s_id not in ( select sc.s_id from score sc inner 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" ) -- 15、查詢兩門及其以上不及格課程的同學的學號，姓名及其平均成績 select st.s_id,st.s_name,avg(sc.s_score) from student st left join score sc on sc.s_id=st.s_id where sc.s_id in ( select sc.s_id from score sc where sc.s_score<60 or sc.s_score is NULL group by sc.s_id having COUNT(sc.s_id)>=2 ) group by st.s_id -- 16、檢索"01"課程分數小於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<60 order 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 st left 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_id group 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：>=90 select 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 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 -- 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 -- 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_id where c.c_id="01" order by sc.s_score desc) c1 , (select @i:=0) a union 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_id where c.c_id="02" order by sc.s_score desc) c2 , (select @ii:=0) aa union all select 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_id where c.c_id="03" order by sc.s_score desc) c3; set @iii=0; -- 20、查詢學生的總成績並進行排名 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 st left join score sc on sc.s_id=st.s_id group 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_id group by t.t_id order 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 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 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 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 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 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 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 st left join score sc on sc.s_id=st.s_id group 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) a union 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) b union all select 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_id inner join student st on st.s_id=c.c_id group 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_id inner 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，並統計同名人數 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 c inner 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、查詢平均成績大於等於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 st left join score sc on sc.s_id=st.s_id group by st.s_id having avg(sc.s_score)>=85 -- 34、查詢課程名稱為"數學"，且分數低於60的學生姓名和分數 select st.s_id,st.s_name,sc.s_score from student st inner join score sc on sc.s_id=st.s_id and sc.s_score<60 inner 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 st left join score sc on sc.s_id=st.s_id left join course c on c.c_id =sc.c_id order 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 st2 left join score sc2 on sc2.s_id=st2.s_id left 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 st inner join score sc on sc.s_id=st.s_id and sc.s_score<60 inner 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 st inner 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 c inner join score sc on sc.c_id=c.c_id group 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 st inner join score sc on sc.s_id=st.s_id inner 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 desc limit 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_id left join course c on c.c_id=sc.c_id where ( select count(1) from student st2 left join score sc2 on sc2.s_id=st2.s_id left join course c2 on c2.c_id=sc2.c_id where 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 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 0,2) a union all select b.* from (select st.s_id,st.s_name,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 0,2) b union all select c.* from (select st.s_id,st.s_name,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 0,2) c -- 借鑒(更准確,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 sc left join course c on c.c_id=sc.c_id group by c.c_id having count(1)>5 order 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_id group 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_id group 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'))