網上流傳的50道sql練習，發現其答案部分有問題，故在此記錄下個人用mysql 的實現。
數據表介紹
--1.學生表
Student(SId,Sname,Sage,Ssex)
--SId 學生編號,Sname 學生姓名,Sage 出生年月,Ssex 學生性別
--2.課程表
Course(CId,Cname,TId)
--CId 課程編號,Cname 課程名稱,TId 教師編號
--3.教師表
Teacher(TId,Tname)
--TId 教師編號,Tname 教師姓名
--4.成績表
SC(SId,CId,score)
--SId 學生編號,CId 課程編號,score 分數
學生表 Student
-- ---------------------------- -- Table structure for `student` -- ---------------------------- DROP TABLE IF EXISTS `student`; CREATE TABLE `student` ( `SId` varchar(10) DEFAULT NULL, `Sname` varchar(10) DEFAULT NULL, `Sage` datetime DEFAULT NULL, `Ssex` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of student -- ---------------------------- INSERT INTO `student` VALUES ('01', '趙雷', '1990-05-18 00:00:00', '男'); INSERT INTO `student` VALUES ('02', '錢電', '1990-05-24 00:00:00', '男'); INSERT INTO `student` VALUES ('03', '孫風', '1990-05-20 00:00:00', '男'); INSERT INTO `student` VALUES ('04', '李雲', '1990-05-25 00:00:00', '男'); INSERT INTO `student` VALUES ('05', '周梅', '1991-12-01 00:00:00', '女'); INSERT INTO `student` VALUES ('06', '吳蘭', '1992-01-01 00:00:00', '女'); INSERT INTO `student` VALUES ('07', '鄭竹', '1989-01-01 00:00:00', '女'); INSERT INTO `student` VALUES ('09', '張三', '2017-12-20 00:00:00', '女'); INSERT INTO `student` VALUES ('10', '李四', '2017-12-25 00:00:00', '女'); INSERT INTO `student` VALUES ('11', '李四', '2012-06-06 00:00:00', '女'); INSERT INTO `student` VALUES ('12', '趙六', '2013-06-13 00:00:00', '女'); INSERT INTO `student` VALUES ('13', '孫七', '2014-06-01 00:00:00', '女');
科目表 Course
-- ---------------------------- -- Table structure for `course` -- ---------------------------- DROP TABLE IF EXISTS `course`; CREATE TABLE `course` ( `CId` varchar(10) DEFAULT NULL, `Cname` varchar(10) DEFAULT NULL, `TId` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of course -- ---------------------------- 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 Teacher
-- ---------------------------- -- Table structure for `teacher` -- ---------------------------- DROP TABLE IF EXISTS `teacher`; CREATE TABLE `teacher` ( `TId` varchar(10) DEFAULT NULL, `Tname` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of teacher -- ---------------------------- INSERT INTO `teacher` VALUES ('01', '張三'); INSERT INTO `teacher` VALUES ('02', '李四'); INSERT INTO `teacher` VALUES ('03', '王五');
成績表 SC
-- ---------------------------- -- Table structure for `sc` -- ---------------------------- DROP TABLE IF EXISTS `sc`; CREATE TABLE `sc` ( `SId` varchar(10) DEFAULT NULL, `CId` varchar(10) DEFAULT NULL, `score` decimal(18,1) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of sc -- ---------------------------- INSERT INTO `sc` VALUES ('01', '02', '90.0'); INSERT INTO `sc` VALUES ('01', '01', '80.0'); INSERT INTO `sc` VALUES ('01', '03', '99.0'); INSERT INTO `sc` VALUES ('02', '02', '60.0'); INSERT INTO `sc` VALUES ('02', '01', '70.0'); INSERT INTO `sc` VALUES ('02', '03', '80.0'); INSERT INTO `sc` VALUES ('03', '01', '80.0'); INSERT INTO `sc` VALUES ('03', '02', '80.0'); INSERT INTO `sc` VALUES ('03', '03', '80.0'); INSERT INTO `sc` VALUES ('04', '01', '50.0'); INSERT INTO `sc` VALUES ('04', '02', '30.0'); INSERT INTO `sc` VALUES ('04', '03', '20.0'); INSERT INTO `sc` VALUES ('05', '01', '76.0'); INSERT INTO `sc` VALUES ('05', '02', '87.0'); INSERT INTO `sc` VALUES ('06', '01', '31.0'); INSERT INTO `sc` VALUES ('06', '03', '34.0'); INSERT INTO `sc` VALUES ('07', '02', '90.0'); INSERT INTO `sc` VALUES ('07', '03', '98.0');
下面是題目和mysql實現
-- 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.*, class1, class2 FROM student st JOIN (SELECT * FROM (SELECT SId s1, score class1 FROM sc WHERE CId = '01') c1, (SELECT SId s2, score class2 FROM sc WHERE CId = '02') c2 WHERE s1 = s2 AND class1 > class2) c ON st.SId = c.s1; -- 1.1 查詢同時存在" 01 "課程和" 02 "課程的情況 SELECT * FROM ((SELECT * FROM sc WHERE CId = '01') t1 JOIN (SELECT * FROM sc WHERE CId = '02') t2 ON t1.Sid = t2.Sid); -- 1.2 查詢存在" 01 "課程但可能不存在" 02 "課程的情況(不存在時顯示為 null ) SELECT * FROM ((SELECT * FROM sc WHERE CId = '01') t1 left JOIN (SELECT * FROM sc WHERE CId = '02') t2 ON t1.Sid = t2.Sid); -- 1.3 查詢不存在" 01 "課程但存在" 02 "課程的情況 select * from sc where sc.CId = '02' AND sc.SId NOT IN (SELECT SId FROM sc WHERE sc.CId = '01'); -- 2.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.SId, Sname, AVG(score) 'avg' FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 60; -- 3.查詢在 SC 表存在成績的學生信息 SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); -- 4.查詢所有同學的學生編號、學生姓名、選課總數、所有課程的成績總和 SELECT st.sid, st.Sname, COUNT(sc.CId) 'nums', SUM(sc.score) 'sum' FROM student st JOIN sc on st.SId = sc.SId GROUP BY st.SId; -- 4.1顯示沒選課的學生(顯示為NULL) SELECT st.sid, st.Sname, CASE WHEN COUNT(sc.SId) > 0 THEN COUNT(sc.SId) ELSE NULL END 'nums', SUM(sc.score) 'sum' FROM student st left JOIN sc on st.SId = sc.SId GROUP BY st.SId; -- 4.2查有成績的學生信息 SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); -- 適用於右表小 SELECT * FROM student st WHERE EXISTS (SELECT * FROM sc WHERE sc.SId = st.SId); -- 適用於右表大 -- 5.查詢「李」姓老師的數量 SELECT COUNT(*) FROM teacher t WHERE t.Tname LIKE '李%'; -- 6.查詢學過「張三」老師授課的同學的信息 SELECT * FROM student st WHERE st.SId IN ( SELECT SId from sc WHERE sc.CId = ( SELECT CId FROM course WHERE course.TId = ( SELECT TId FROM teacher WHERE teacher.Tname = '張三'))); -- 7.查詢沒有學全所有課程的同學的信息 SELECT * FROM student WHERE SId NOT IN( SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course)); -- 8.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 * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01')); -- 9.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 -- 解法一 SELECT * FROM student WHERE SId IN (SELECT SId FROM sc GROUP BY SId HAVING SId <> '01' AND GROUP_CONCAT(CId ORDER BY CId) = (SELECT GROUP_CONCAT(CId ORDER BY CId) FROM sc WHERE SId = '01')); -- 解法二 SELECT * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01') AND SId <> '01' GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM sc WHERE SId = '01')); -- 10.查詢沒學過"張三"老師講授的任一門課程的學生姓名 SELECT student.Sname FROM student WHERE SId NOT IN (SELECT SId FROM sc WHERE CId in (SELECT course.CId FROM course WHERE TId = (SELECT teacher.TId FROM teacher WHERE tname = '張三'))); -- 11.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.SId, sname, avg(score) FROM student st JOIN sc ON st.sid = sc.SId GROUP BY st.SId HAVING COUNT(score <= 60 OR NULL) >= 2; -- 錯誤解法： select student.sid, student.sname, AVG(sc.score) from student,sc where student.sid = sc.sid and sc.score< 60 group by sc.sid having count(*)>1; -- 12.檢索" 01 "課程分數小於 60，按分數降序排列的學生信息 SELECT st.*, score FROM student st JOIN sc ON st.SId = sc.SId AND sc.CId = '01' AND sc.score < 60 ORDER BY score DESC; -- 13.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 SELECT * from sc JOIN (SELECT SId, avg(score) avg FROM sc GROUP BY SId) avgs ON sc.SId = avgs.SId ORDER BY avgs.avg desc; -- 14.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 -- 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.CId '課程 ID', count(*) '選修人數', course.Cname '課程 name', MAX(score) '最高分', MIN(score) '最低分', AVG(score) '平均分', COUNT(score >= 60 OR NULL) / COUNT(*) '及格率', COUNT(score >= 70 AND score < 80 OR NULL) / COUNT(*) '中等率', COUNT(score >= 80 AND score < 90 OR NULL) / COUNT(*) '優良率', COUNT(score >= 90 OR NULL) / COUNT(*) '優秀率' FROM sc join course ON sc.CId = course.CId GROUP BY sc.CId ORDER BY COUNT(*) desc, sc.CId; -- 15.按各科成績進行排序，並顯示排名， Score 重復時保留名次空缺 SELECT CId, SId, score , (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 rank FROM sc o ORDER BY CId, score DESC; -- 16.查詢學生的總成績，並進行排名，總分重復時不保留名次空缺 SET @rank = 0; SELECT q.sid, q.sum, (@rank := @rank + 1) rank FROM (SELECT sid, sum(score) sum FROM sc GROUP BY SId ORDER BY sum) q; -- 17.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]，[60-0] 及所占百分比 SELECT s.CId, c.Cname, CONCAT(COUNT(score >= 85 OR NULL), ', ',COUNT(score >= 85 OR NULL) / COUNT(*)) '[100-85]', COUNT(score < 85 AND score >= 70 OR NULL) '[85-70]', COUNT(score < 70 AND score >= 60 OR NULL) '[70-60]', COUNT(score < 60 OR NULL) '[60-0]' FROM sc s JOIN course c ON s.CId = c.CId GROUP BY s.CId; -- 18. 查詢各科成績前三名的記錄 SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE o.CId = CId AND o.score < score) + 1 <= 3 ORDER BY CId, score DESC, SId; -- 19.查詢每門課程被選修的學生數 SELECT CId, COUNT(*) FROM sc GROUP BY CId; -- 20.查詢出只選修兩門課程的學生學號和姓名 SELECT st.SId, st.Sname FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING COUNT(*) = 2; -- 21.查詢男生、女生人數 select ssex, count(*) from student group by ssex; -- 22.查詢名字中含有「風」字的學生信息 SELECT * FROM student WHERE Sname LIKE '%風%'; -- 23.查詢同名學生名單，並統計同名人數 SELECT st.*, (SELECT COUNT(*) FROM student WHERE Sname = st.Sname) 'same' FROM student st WHERE st.SId IN (SELECT s1.SId FROM student s1 JOIN student s2 ON s1.SId != s2.SId AND s1.Sname = s2.Sname); -- 24.查詢 1990 年出生的學生名單 SELECT * FROM student WHERE YEAR(Sage) = 1990; -- 25.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 CId, avg(score) avg FROM sc GROUP BY CId ORDER BY avg DESC, CId; -- 26.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.SId, Sname, AVG(score) avg FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 85; -- 27.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.Sname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.CId = c.CId AND s.score < 60 AND c.Cname = '數學'; -- 28.查詢所有學生的課程及分數情況（存在學生沒成績，沒選課的情況） SELECT Sname, st.SId, score FROM student st LEFT JOIN sc s ON st.SId = s.SId; -- 29.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 -- 這個我沒太理解，理解一是任意一門成績均在70分以上 SELECT st.Sname, c.Cname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN (SELECT sid FROM sc GROUP BY SId HAVING MIN(score) > 70); -- 理解二是存在一門成績在70分以上即可滿足條件 SELECT st.Sname, c.Cname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN (SELECT sid FROM sc GROUP BY SId HAVING MAX(score) > 70); -- 理解三就是找出所有大於70分的得分。 select student.sname, course.cname,sc.score from student,course,sc where sc.score>70 and student.sid = sc.sid and sc.cid = course.cid; SELECT * FROM sc WHERE score > 70 -- 30.查詢存在不及格的課程 SELECT cid FROM sc GROUP BY CId HAVING MIN(score) < 60; select cid from sc where score< 60 group by cid; SELECT DISTINCT cid FROM sc WHERE score < 60; -- 31.查詢課程編號為 01 且課程成績在 80 分及以上的學生的學號和姓名 SELECT sid, sname FROM student WHERE sid IN (SELECT SId FROM sc WHERE CId = '01' AND score >= 80); -- 32.求每門課程的學生人數 SELECT cid, count(*) FROM sc GROUP BY cid; -- 33.成績不重復，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.*, score FROM student st JOIN sc s ON st.SId = s.SId AND s.CId = (SELECT CId FROM course WHERE TId = (SELECT TId FROM teacher WHERE Tname = '張三')) ORDER BY score DESC LIMIT 1; -- 34.成績有重復的情況下，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.*, score, CId FROM student st JOIN sc s ON st.SId = s.SId AND s.CId = (SELECT CId FROM course WHERE TId = (SELECT TId FROM teacher WHERE Tname = '張三')) WHERE score = (SELECT max(score) FROM sc WHERE CId = s.CId); -- 35.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 -- 這個問題其實一開始沒太明白啥意思，后來理解為某個人的幾科分數是一樣的，需要把這個人找出來 select a.cid, a.sid, a.score from sc as a inner join sc as b on a.sid = b.sid and a.cid != b.cid and a.score = b.score group by cid, sid; -- 36.查詢每門功成績最好的前兩名 SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 <= 2 ORDER BY o.cid, sid; -- 37.統計每門課程的學生選修人數（超過 5 人的課程才統計） SELECT CId, COUNT(*) sum FROM sc GROUP BY CId HAVING sum > 5; -- 38.檢索至少選修兩門課程的學生學號 SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) >= 2; -- 39.查詢選修了全部課程的學生信息 SELECT * FROM student WHERE sid IN (SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course)); -- 40.查詢各學生的年齡，只按年份來算 SELECT SId, Sname, Sage, YEAR(NOW()) - YEAR(Sage) FROM student; -- 41.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 SId '學生編號', Sname '學生姓名', TIMESTAMPDIFF(YEAR, Sage, NOW()) '學生年齡' FROM student; -- 42.查詢本周過生日的學生 -- 有點復雜，需要拼接出本周的起止日期 SELECT * FROM student WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage))) BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) DAY)) AND DATE(DATE_ADD(NOW(),INTERVAL 6 - WEEKDAY(NOW()) DAY)); SELECT * FROM student -- 43. 查詢下周過生日的學生 -- 同42 SELECT * FROM student WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage))) BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) - 7 DAY)) AND DATE(DATE_ADD(NOW(),INTERVAL 13 - WEEKDAY(NOW()) DAY)); -- 44.查詢本月過生日的學生 SELECT * FROM student WHERE month(sage) = month(NOW()) -- 45.查詢下月過生日的學生 -- 注意本月是12月的話，下一個月份是1即可 SELECT * FROM student WHERE month(Sage) = (CASE WHEN month(NOW()) = 12 THEN 1 ELSE MONTH(NOW()) + 1 END);