根据上一篇文章建立的表,我们来做一些多表练习: 没建立表的可以点击此链接去建立练习用的表: 目录 1.查询“1”号学生的姓名和各科成绩: 2.查询各个学科的平均成绩和最高成绩: 3.查询所有姓张的同学的各科成绩: 4.查询每个同学的最高成
根据上一篇文章建立的表,我们来做一些多表练习:
没建立表的可以点击此链接去建立练习用的表:
目录
12.查询被"Tom"和"Jerry"教的课程的最高分和最低分
15.查询课程编号为1且课程成绩在60分以上的学生的学号和姓名(子查询)
16. 查询平均成绩大于等于70的所有学生学号、姓名和平均成绩
19.查询每门课程的平均成绩,结果按照平均成绩降序排列,如果平均成绩相同,再按照课程编号升序排列
20.查询平均成绩大于60分的同学的学生编号和学生姓名和平均成绩
28.查询课程名称为"java",且分数低于60分的学生姓名和分数
进行student表和scores表的id相连接,course表和scores表的id相连接
SELECTs.id sid,s.`name` sname,c.`name` cname,sc.score FROMstudent sLEFT JOIN scores sc ON s.id = sc.s_idLEFT JOIN course c ON c.id = sc.c_id WHEREs.id = 1;
SELECTc.id,c.`name`,AVG( sc.score ),max( sc.score ) FROMcourse cLEFT JOIN scores sc ON c.id = sc.c_id GROUP BYc.id,c.`name`;
SELECTs.id,s.`name`,c.`name` cname,sc.score FROMstudent sLEFT JOIN scores sc ON sc.s_id = s.idLEFT JOIN course c ON c.id = sc.c_id WHEREs.`name` LIKE '张%';
SELECTt.id,t.NAME,c.id,c.NAME,r.score FROM(SELECTs.id,s.NAME,(SELECTmax( score ) FROMscores r WHEREr.s_id = s.id ) score FROMstudent s ) tLEFT JOIN scores r ON r.s_id = t.id AND r.score = t.scoreLEFT JOIN course c ON r.c_id = c.id;
SELECT* FROMstudent s WHEREid IN (SELECT DISTINCTr.s_id FROM(SELECTc.id,c.NAME,max( score ) score FROMstudent sLEFT JOIN scores r ON r.s_id = s.idLEFT JOIN course c ON c.id = r.c_id GROUP BYc.id,c.NAME ) tLEFT JOIN scores r ON r.c_id = t.id AND t.score = r.score );
SELECTs.id,s.NAME sname,sc.score,c.NAME FROMstudent sLEFT JOIN scores sc ON s.id = sc.s_idLEFT JOIN course c ON sc.c_id = c.id WHEREs.NAME LIKE '%张%' OR s.NAME LIKE '%李%';
SELECT* FROMstudent WHEREid IN (SELECTsc.s_id FROMscores sc GROUP BYsc.s_id HAVINGavg( sc.score ) >= 70 );
SELECTs.id,s.NAME,sum( sc.score ) score FROMstudent sLEFT JOIN scores sc ON s.id = sc.s_id GROUP BYs.id,s.NAME ORDER BYscore DESC,s.id ASC;
SELECTc.NAME,max( sc.score ),min( sc.score ),avg( sc.score ) FROMcourse cLEFT JOIN scores sc ON c.id = sc.c_id WHEREc.NAME = '数学';
SELECTc.id,c.NAME,avg( sc.score ) score FROMcourse cLEFT JOIN scores sc ON c.id = sc.c_id GROUP BYc.id,c.NAME ORDER BYscore DESC;
SELECTt.id,t.NAME,c.id cid,c.NAME cname,avg( r.score ) FROMteacher tLEFT JOIN course c ON t.id = c.t_idLEFT JOIN scores r ON r.c_id = c.id GROUP BYt.id,t.NAME,c.id,c.NAME;
SELECTt.id,t.NAME,c.id cid,c.NAME cname,max( r.score ),min( r.score ) FROMteacher tLEFT JOIN course c ON t.id = c.t_idLEFT JOIN scores r ON r.c_id = c.id GROUP BYt.id,t.NAME,c.id,c.NAME HAVINGt.NAME IN ( 'Tom', 'Jerry' );
SELECTt.id,t.sname,r.c_id,c.NAME,t.score FROM(SELECTs.id,s.NAME sname,max( r.score ) score FROMstudent sLEFT JOIN scores r ON r.s_id = s.id GROUP BYs.id,s.NAME ) tLEFT JOIN scores r ON r.s_id = t.id AND r.score = t.scoreLEFT JOIN course c ON r.c_id = c.id;
SELECTs.id,s.NAME,c.id,c.NAME,r.score FROMstudent sLEFT JOIN scores r ON s.id = r.s_idLEFT JOIN course c ON c.id = r.c_id;
SELECTs.*,r.* FROMstudent sLEFT JOIN scores r ON s.id = r.s_id WHEREr.c_id = 1 AND r.score > 60
SELECTs.id,s.NAME,t.score FROMstudent sLEFT JOIN ( SELECT r.s_id, avg( r.score ) score FROM scores r GROUP BY r.s_id ) t ON s.id = t.s_id WHEREt.score >= 70;
SELECT* FROMstudent s WHEREid IN ( SELECT r.s_id FROM scores r GROUP BY r.s_id HAVING min( r.score ) < 60 );
SELECTc.id,c.NAME,count(*) FROMcourse cLEFT JOIN scores r ON c.id = r.c_id GROUP BYc.id,c.NAME;
SELECTc.id,c.NAME,avg( score ) score FROMcourse cLEFT JOIN scores r ON c.id = r.c_id GROUP BYc.id,c.NAME ORDER BYscore DESC,c.id ASC;
SELECTs.id,s.NAME sname,avg( r.score ) score FROMstudent sLEFT JOIN scores r ON r.s_id = s.idLEFT JOIN course c ON c.id = r.c_id GROUP BYs.id,s.NAME HAVINGscore > 65;
SELECTs.id,s.NAME,s.gender FROMstudent sLEFT JOIN scores r ON s.id = r.s_id WHEREr.score > 80 GROUP BYs.id,s.NAME,s.gender HAVINGcount(*) = 1;
SELECTs.id,s.NAME,s.gender FROMstudent sLEFT JOIN scores r ON s.id = r.s_id GROUP BYs.id,s.NAME,s.gender HAVINGcount(*) = 3;
SELECT* FROMcourse c WHEREid IN (SELECTr.c_id FROMscores r GROUP BYr.c_id HAVINGmin( r.score ) < 60 );
SELECTs.id,s.NAME FROMstudent sLEFT JOIN scores r ON s.id = r.s_id GROUP BYs.id,s.NAME HAVINGcount(*) >= 4;
SELECT* FROMstudent WHEREid IN (SELECTr.s_id FROMscores r GROUP BYr.s_id HAVINGcount(*) != 5);
SELECTs.id,s.NAME,count(*) number FROMstudent sLEFT JOIN scores r ON s.id = r.s_id GROUP BYs.id,s.NAME HAVINGnumber = ( SELECT count(*) FROM course );
SELECTs.id,s.NAME,count(*) number FROMstudent sLEFT JOIN scores r ON s.id = r.s_id GROUP BYs.id,s.NAME;
SELECTs.id,s.NAME,r.score FROMstudent sLEFT JOIN scores r ON s.id = r.s_idLEFT JOIN course c ON r.c_id = c.id WHEREc.NAME = 'java' AND r.score < 60;
SELECTs.id,s.NAME FROMstudent sLEFT JOIN scores r ON r.s_id = s.idLEFT JOIN course c ON c.id = r.c_idLEFT JOIN teacher t ON t.id = c.t_id WHEREt.NAME = 'Tom';
SELECT* FROMstudent WHEREid NOT IN (SELECT DISTINCTs.id FROMstudent sLEFT JOIN scores r ON r.s_id = s.idLEFT JOIN course c ON c.id = r.c_idLEFT JOIN teacher t ON t.id = c.t_id WHEREt.NAME = 'Tom' )
来源地址:https://blog.csdn.net/weixin_49627122/article/details/126380916
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