MYSQL题目练习专用
1.表名称和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
2.插入测试数据
use test drop table sc,student,teacher 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`) ); --课程表 CREATE TABLE `Course`( `c_id` VARCHAR(20), `c_name` VARCHAR(20) NOT NULL DEFAULT '', `t_id` VARCHAR(20) NOT NULL, PRIMARY KEY(`c_id`) ); --教师表 CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(`t_id`) ); --成绩表 CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`) ); --插入学生表测试数据 insert into Student values('01' , '赵雷' , '1990-01-01' , '男'); insert into Student values('02' , '钱电' , '1990-12-21' , '男'); insert into Student values('03' , '孙风' , '1990-05-20' , '男'); insert into Student values('04' , '李云' , '1990-08-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '女'); insert into Student values('06' , '吴兰' , '1992-03-01' , '女'); insert into Student values('07' , '郑竹' , '1989-07-01' , '女'); insert into Student values('08' , '王菊' , '1990-01-20' , '女'); --课程表测试数据 insert into Course values('01' , '语文' , '02'); insert into Course values('02' , '数学' , '01'); insert into Course values('03' , '英语' , '03'); --教师表测试数据 insert into Teacher values('01' , '张三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五'); --成绩表测试数据 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);
准备条件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否则此种情况下会报错:
-- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo.
-- 原因:
-- MySQL 5.7.5和up实现了对功能依赖的检测。如果启用了only_full_group_by SQL模式(在默认情况下是这样),
-- 那么MySQL就会拒绝选择列表、条件或顺序列表引用的查询,这些查询将引用组中未命名的非聚合列,而不是在功能上依赖于它们。
-- (在5.7.5之前,MySQL没有检测到功能依赖项,only_full_group_by在默认情况下是不启用的。关于前5.7.5行为的描述,请参阅MySQL 5.6参考手册。)
-- 执行以下个命令,可以查看 sql_mode 的内容。
-------------------------------下边是重点
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';
3.练习题目
我这是5题一小节
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.*,c.c_name,sc.s_score FROM student st join score sc on st.s_id = sc.s_id join course c on sc.c_id = c.c_id where (SELECT s_score FROM score c_id="01") > (SELECT s_score FROM score c_id="02") 01,02---course,score s_score---score student.*---student --答案: SELECT st.*,sc1.s_score,sc2.s_score from student st LEFT JOIN score sc1 ON sc1.s_id = st.s_id AND sc1.c_id = "01" LEFT JOIN score sc2 ON sc2.s_id = st.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 --模仿着第一题写出来的 select st.*,sc1.s_score as '数学',sc2.s_score '语文' from student st LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.c_id = '01' LEFT JOIN score sc2 on sc2.s_id = st.s_id and sc2.c_id = '02' where sc1.s_score < sc2.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*** --自己憋出来的答案 select st.s_id,st.s_name,SUM(sc.s_score),ROUND(AVG(sc.s_score),3) from (student st) JOIN (score sc) ON sc.s_id = st.s_id GROUP BY st.s_id HAVING AVG(sc.s_score) >= 60 --答案 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,ROUND(AVG(sc.s_score),2) '平均成绩' 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 --答案 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(DISTINCT c.c_id),SUM(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 GROUP BY st.s_id --答案 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、查询"李"姓老师的数量 SELECT COUNT(t_id) from teacher where 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 join score sc on sc.s_id = st.s_id join course c on c.c_id = sc.c_id join teacher t on t.t_id = c.t_id where t_name = "张三" --答案 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="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 --琢磨结果 select * from student where s_id not in (select st.s_id from student st join score sc on sc.s_id = st.s_id JOIN course c on c.c_id = sc.c_id join teacher t on t.t_id = c.t_id WHERE t.t_name = "张三") --答案 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="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 SELECT st.* from student st where st.s_id in (SELECT sc1.s_id from score sc1 where sc1.c_id = "01" and sc1.s_id in (SELECT sc2.s_id from score sc2 where sc2.c_id = "02")) --答案 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" ) --高手 SELECT st.* FROM student st INNER 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 WHERE st.s_id in (select sc1.s_id from score sc1 where sc1.c_id = "01" and sc1.s_id not in (select sc2.s_id from score sc2 where sc2.c_id = "02")) --答案 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、查询没有学全所有课程的同学的信息 select st.* from student st LEFT JOIN score sc on st.s_id = sc.s_id LEFT JOIN course c on c.c_id = sc.c_id GROUP BY st.s_id HAVING COUNT(c.c_id) < (SELECT COUNT(1) from course) --答案--居然和我的类似(我服了你了~) select st.* from Student st left join Score S on st.s_id = S.s_id group by st.s_id having count(c_id)<(select count(c_id) 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 DISTINCT st.* from student st LEFT JOIN score sc on sc.s_id = st.s_id where sc.c_id in (select sc.c_id from score sc where sc.s_id = '01') --答案(比较臃肿) 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、查询和"01"号的同学学习的课程完全相同的其他同学的信息 -- CONCAT_WS(separator,str1,str2,...)的解法,有点臃肿 select st.*,CONCAT_WS(':',sc1.c_id,sc2.c_id,sc3.c_id) as pigpigpigpigpigpig from student st LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.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' GROUP BY st.s_id HAVING pigpigpigpigpigpig in (select CONCAT_WS(':',sc1.c_id,sc2.c_id,sc3.c_id) as xxx from student st LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.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' where st.s_id = '01') select s_id from score GROUP BY s_id select s_id,GROUP_CONCAT(c_id SEPARATOR ';')from score GROUP BY s_id HAVING s_id = '01' -- 重要的一步,,意思:把s_id对应的c_id用分隔符':'连接起来成为新的字段 --我的答案 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 SEPARATOR ':') = (select GROUP_CONCAT(c_id SEPARATOR ':') from score GROUP BY s_id HAVING s_id = '01') --答案 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、查询没学过"张三"老师讲授的任一门课程的学生姓名 select DISTINCT st.s_name 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 c.c_id in (select c.c_id from course c join teacher t on t.t_id = c.t_id where t.t_name = '张三') --答案 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="张三" ) 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 --重要,因为我没做出来 select st.s_id,st.s_name,AVG(sc.s_score) as '平均成绩' 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 GROUP BY sc.s_id HAVING COUNT(sc.s_id) >= 2) GROUP BY st.s_id --答案 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、检索"01"课程分数小于60,按分数降序排列的学生信息 select st.* from student st join score sc on sc.s_id = st.s_id WHERE sc.c_id = '01' and sc.s_score < 60 ORDER BY sc.s_score DESC --答案 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.s_id as '学生编号',st.s_name,sc1.s_score as '语文',sc2.s_score as '数学',sc3.s_score as '外语',ROUND(AVG(sc4.s_score),2) as '平均成绩' from student st LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.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 --答案 select st.s_id,st.s_name,avg(sc4.s_score) "平均分",sc.s_score "语文",sc2.s_score "数学",sc3.s_score "英语" 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 --我的答案有误 select c.c_id as '课程ID',c.c_name as '课程name',MAX(sc.s_score) as '最高分',MIN(sc.s_score) as '最低分',AVG(sc.s_score) as '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) as '及格率', COUNT(sc.s_score >=70 and sc.s_score < 80)/COUNT(sc.s_score) as '中等率',COUNT(sc.s_score >=80 and sc.s_score < 90)/COUNT(sc.s_score) as '优良率',COUNT(sc.s_score >= 90)/COUNT(sc.s_score) as '优秀率' from score sc LEFT JOIN course c on c.c_id = sc.c_id GROUP BY c.c_id --答案 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)) "优良率" ,((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)) "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 19、按各科成绩进行排序,并显示排名(实现不完全) ***多看多练 -- mysql没有rank函数 select c1.s_id,c1.c_id,c1.c_name,@score := c1.s_score,@x := @x + 1 from (SELECT sc.s_id,c.c_name,c.c_id,sc.s_score from course c join score sc on sc.c_id = c.c_id where c.c_id = '01' ORDER BY sc.s_score DESC) as c1, (select @x := 0) y UNION ALL (select c2.s_id,c2.c_id,c2.c_name,@score := c2.s_score,@q := @q + 1 from (SELECT sc.*,c.c_name 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) as c2, (select @q := 0) w) UNION ALL (select c3.s_id,c3.c_id,c3.c_name,@score := c3.s_score,@e := @e + 1 from (select sc.*,c.c_name 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)as c3, (select @e := 0) r) --验证: select COUNT(1) as '各科课程数量' from score GROUP BY c_id --答案 -- 加@score是为了防止用union all 后打乱了顺序 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、查询学生的总成绩并进行排名 --改进 select s1.s_id,s1.total,@i := @i + 1 as '排名' from (select st.s_id,(case when SUM(sc.s_score) is null then 0 else SUM(sc.s_score) end) as total 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) as s1, (select @i := 0) j --答案 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、查询不同老师所教不同课程平均分从高到低显示 select t.t_name,t.t_id,c.c_name,ROUND(AVG(sc.s_score),2) as '平均成绩' from score sc LEFT JOIN course c on c.c_id = sc.c_id LEFT JOIN teacher t on t.t_id = c.t_id GROUP BY t.t_id ORDER BY AVG(sc.s_score) DESC --答案 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 st.*,kkk.* FROM (select c1.s_id,c1.c_name,@score := c1.s_score,@x := @x + 1 as Rank from (select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc LEFT JOIN course c on c.c_id = sc.c_id where c.c_id = '01' ORDER BY sc.s_score DESC LIMIT 1,2) as c1, (SELECT @x := 1) y UNION ALL (select c2.s_id,c2.c_name,@score := c2.s_score,@xx := @xx + 1 as Rank from (select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc LEFT JOIN course c on c.c_id = sc.c_id where c.c_id = '02' ORDER BY sc.s_score DESC LIMIT 1,2) as c2, (SELECT @xx := 1) yy) UNION ALL (select c3.s_id,c3.c_name,@score := c3.s_score,@xxx := @xxx + 1 as Rank from (select sc.s_id,c.c_name,c.c_id,sc.s_score from score sc LEFT JOIN course c on c.c_id = sc.c_id where c.c_id = '03' ORDER BY sc.s_score DESC LIMIT 1,2) as c3, (SELECT @xxx := 1) yyy)) as kkk LEFT JOIN student st on st.s_id = kkk.s_id --答案 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、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 sc where sc.c_id=c.c_id and sc.s_score <= 100 and sc.s_score >= 85)/(select COUNT(1) from score sc where c.c_id = sc.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 c.c_id = sc.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 c.c_id = sc.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 c.c_id = sc.c_id)) "60-0" from course c ORDER BY c.c_id ASC --答案 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、查询学生平均成绩及其名次 select s1.s_id,s1.s_name,@xxx := s1.kkk as '平均成绩',@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) as kkk FROM student st LEFT JOIN score sc on st.s_id = sc.s_id GROUP BY st.s_id ORDER BY AVG(sc.s_score) DESC) as s1, (select @i := 0) j --重要的语句 case when AVG(sc.s_score) is null then 0 else AVG(sc.s_score) end --答案 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、查询各科成绩前三名的记录 select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i := @i + 1 from (select st.s_id,st.s_name,sc.c_id,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 where sc.c_id = '01' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, (select @i := 0) as j UNION ALL (select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i2 := @i2 + 1 from (select st.s_id,st.s_name,sc.c_id,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 where sc.c_id = '02' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, (select @i2 := 0) as j2 ) UNION ALL (select s1.s_id,s1.s_name,s1.c_id,s1.c_name,s1.s_score,@i3 := @i3 + 1 from (select st.s_id,st.s_name,sc.c_id,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 where sc.c_id = '03' GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,3) as s1, (select @i3 := 0) as j3 ) --答案 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、查询每门课程被选修的学生数 select sc.c_id,c.c_name,COUNT(1) as '选修人数' from score sc LEFT JOIN course c on c.c_id = sc.c_id GROUP BY sc.c_id --验证:select COUNT(s_score) FROM score --答案 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、查询出只有两门课程的全部学生的学号和姓名 select st.s_id,st.s_name from student st LEFT JOIN score sc on sc.s_id = st.s_id GROUP BY st.s_id HAVING COUNT(1) = 2 --答案 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、查询男生、女生人数 select s_sex as '性别',COUNT(1) as '人数' from student st GROUP BY s_sex --答案 select st.s_sex,count(1) from student st group by st.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%' --答案 select st.* from student st where st.s_name like "%风%"; 30、查询同名同性学生名单,并统计同名人数 select st.s_name,COUNT(1) from student st GROUP BY st.s_name,st.s_sex HAVING COUNT(1) >= 2 --答案 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.s_name,st.s_birth from student st where st.s_birth like '1990%' --答案 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 sc.c_id,c.c_name,ROUND(AVG(sc.s_score),2) as '平均成绩' from score sc LEFT JOIN course c on c.c_id = sc.c_id GROUP BY sc.c_id ORDER BY AVG(sc.s_score) DESC ,sc.c_id ASC --答案 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.s_id,st.s_name,ROUND(AVG(sc.s_score),2) as '平均成绩' 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 --答案 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、查询课程名称为"数学",且分数低于60的学生姓名和分数 select st.s_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 where c.c_name = '数学' and sc.s_score < 60 --答案 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 ="数学" 35、查询所有学生的课程及分数情况; select st.s_id,st.s_name,sc.c_id,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,sc.c_id --答案 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.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 where st.s_id in (select sc.s_id FROM score sc where sc.s_score > 70 GROUP BY sc.s_id HAVING COUNT(sc.s_score) = 3) ORDER BY st.s_id ASC --答案,,我的答案和答案有点出入 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、查询不及格的课程 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 where sc.s_score <60 --答案 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.s_id,st.s_name,sc.c_id,sc.s_score from student st JOIN score sc on sc.s_id = st.s_id where sc.c_id = '01' and sc.s_score >= 80 --答案 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、求每门课程的学生人数 select sc.c_id,c.c_name,COUNT(1) from score sc LEFT JOIN course c on c.c_id = sc.c_id GROUP BY sc.c_id --答案 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.*,c1.c_name,c1.kkk as '分数' from student st join (select sc.s_id,c.c_name,MAX(sc.s_score) as kkk from score sc 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 = '张三') as c1 on c1.s_id = st.s_id --答案 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="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 --理解偏题了 select st.s_id,sc1.s_score,sc2.s_score,sc3.s_score from student st LEFT JOIN score sc1 on sc1.s_id = st.s_id and sc1.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' where sc1.s_score != sc2.s_score and sc1.s_score != sc3.s_score and sc2.s_score != sc3.s_score --参考答案 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 --答案 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、查询每门功成绩最好的前两名 select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i :=@i + 1 from (select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '01' LEFT JOIN course c on c.c_id = sc.c_id GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, (select @i := 0) as j UNION ALL (select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i2 :=@i2 + 1 from (select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '02' LEFT JOIN course c on c.c_id = sc.c_id GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, (select @i2 := 0) as j2 ) UNION ALL (select x1.c_id,x1.c_name,x1.s_id,x1.s_name,x1.s_score,@i3 :=@i3 + 1 from (select sc.c_id,c.c_name,st.s_id,st.s_name,sc.s_score from student st LEFT JOIN score sc on sc.s_id = st.s_id and sc.c_id = '03' LEFT JOIN course c on c.c_id = sc.c_id GROUP BY st.s_id ORDER BY sc.s_score DESC LIMIT 0,2) as x1, (select @i3 := 0) as j3 ) --验证: SELECT sc.c_id,sc.s_score from score sc where sc.c_id = '01' ORDER BY sc.s_score desc LIMIT 2 --答案 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 --高手 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 sc.c_id,c.c_name,COUNT(1) from score sc LEFT JOIN course c on c.c_id = sc.c_id GROUP BY sc.c_id HAVING COUNT(1)> 5 ORDER BY COUNT(1) DESC,sc.c_id ASC --答案 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、检索至少选修两门课程的学生学号 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 --答案 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、查询选修了全部课程的学生信息 --借鉴了答案的用法。在最后一句 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 GROUP BY st.s_id HAVING COUNT(1) = (select COUNT(1) from course) --答案 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、查询各学生的年龄 --尝试用函数进行解答,方便快捷,无纰漏 SELECT st.*,TIMESTAMPDIFF(YEAR, st.s_birth, CURDATE()) from student st --答案 select st.*,timestampdiff(year,st.s_birth,now()) from student st 47、查询本周过生日的学生 --这个我的答案有问题 select * from student st where year(st.s_birth)<year(CURDATE()) and month(st.s_birth)=month(CURDATE()) and day(st.s_birth)>day(CURDATE())-DATE_FORMAT(CURDATE()+1,'%e') and day(st.s_birth)<=day(CURDATE())+(8-DATE_FORMAT(CURDATE(),'%e')) --答案 select st.* from student st where week(now())=week(date_format(st.s_birth,'%Y%m%d')) 48、查询下周过生日的学生 我参考的答案 --答案 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.* from student st where month(now()) = month(DATE_FORMAT(st.s_birth,'%Y%m%d')) --答案 select st.* 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 --参考了答案的 select st.* from student st where (month(now())+1) mod 12 = month(DATE_FORMAT(st.s_birth,'%Y%m%d')) --答案; -- 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.* 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'))