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sql题
下面和大家分享一个比较经典的场景，数据库的面试题目，主要的表是学生、课程、成绩、教师四张表，本示例的特点是有模拟数据，加深理解和印象，答案主要基于Oracle来实现的
1、四张表分别为：
Student(S#,Sname,Sage,Ssex) 学生表 S#：学号；Sname：学生姓名；Sage：学生年龄；Ssex：学生性别 Course(C#,Cname,T#) 课程表 C#：课程编号；Cname：课程名字；T#：教师编号 SC(S#,C#,score) 成绩表 S#：学号；C#,课程编号；score：成绩 Teacher(T#,Tname) 教师表 T#：教师编号； Tname：教师名字
2、题目要求
1、查询“111”课程比“112”课程成绩高的所有学生的学号； 2、查询平均成绩大于60分的同学的学号和平均成绩； 3、查询所有同学的学号、姓名、选课数、总成绩； 4、查询姓“李”的老师的个数； 5、查询没学过‘陈奕迅‘老师课的同学的学号、姓名； 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名； 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名； 8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名 9、查询所有课程成绩小于60分的同学的学号、姓名 10、查询没有学全所有课的同学的学号、姓名 11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名 12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名 13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩 14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名 15、删除学习‘陈奕迅‘老师课的SC表记录 16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2号课的平均成绩 17、按平均成绩从高到低显示所有学生的“语文1”、“生物2”、“化学1”三门的课程成绩，按如下形式显示 18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分 19、按各科平均成绩从低到高和及格率的百分数从高到低顺序 20、查询如下课程平均成绩和及格率的百分数(用"1行"显示) 21、查询不同老师所教不同课程平均分从高到低显示 22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：语文1（111），语文2（112），数学1 （113），数学2（114） 23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85] 优秀人数,[85-70] 良好人数,[70-60] 一般人数,[ <60] 刚及格人数 24、查询学生平均成绩及其名次 25、查询各科成绩前三名的记录:(不考虑成绩并列情况) 26、查询每门课程被选修的学生数 27、查询出只选修了一门课程的全部学生的学号和姓名 28、查询男生、女生人数 29、查询名字中有‘黑‘的学生名单 30、查询同名同性学生名单，并统计同名人数
32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列 33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩 34、查询课程名称为‘语文1‘，且分数低于60的学生姓名和分数 35、查询所有学生的选课情况 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 37、查询不及格的课程，并按课程号从大到小排列 38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名 39、求选了课程的学生人数 40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩 41、查询各个课程及相应的选修人数 42、查询不同课程成绩相同的学生的学号、课程号、学生成绩 43、查询每门功成绩最好的前两名 44、统计每门课程的学生选修人数（超过10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，查询结果按人数降序排列，若人数相同，按课程号升序排列 45、检索至少选修两门课程的学生学号 46、查询全部学生都选修的课程的课程号和课程名 47、查询没学过‘陈奕迅‘老师讲授的任一门课程的学生姓名 48、查询两门以上不及格课程的同学的学号及其平均成绩 49、检索‘114‘课程分数小于60，按分数降序排列的同学学号 50、删除‘2‘同学的‘111‘课程的成绩
/*====================================*/
/*=========== 习题开始 =============*/
/*====================================*/
--1、查询“111”课程比“112”课程成绩高的所有学生的学号；
/* EXISTS 版本 */
SELECT t1.s#
FROM sc t1
WHERE t1.c# = 111
AND EXISTS (
SELECT 1
FROM sc t2
WHERE t2.c# = 112
AND t1.s# = t2.s# --要求同一个学生 所以有t1.s# = t2.s#
AND t1.score > t2.score
);
/* 子查询版本 */
SELECT a.s#
FROM (SELECT t1.s#,t1.score FROM sc t1 WHERE t1.c# = 111) a
INNER JOIN (SELECT t2.s#,t2.score FROM sc t2 WHERE t2.c# = 112) b
ON a.s# = b.s#
WHERE a.score > b.score;
--2、查询平均成绩大于60分的同学的学号和平均成绩；
SELECT t1.s# AS ST_CODE,
AVG(t1.score) AS CU_AVG
FROM sc t1
GROUP BY t1.s#
HAVING AVG(t1.score) > 60
--3、查询所有同学的学号、姓名、选课数、总成绩；
/* group by 的时候要注意选出来的字段是不是都是聚合函数或者分组的字段 */
SELECT t1.s# AS ST_DOCE,
MIN(t1.sname) AS ST_NAME,
COUNT(t2.c#) AS CU_NUM,
SUM(t2.score) AS CU_SUM
FROM STUDENT t1
LEFT JOIN SC t2
ON t1.s# = t2.s#
GROUP BY t1.s#
--4、查询姓“李”的老师的个数；
SELECT COUNT(t1.t#) AS 李_NUM
FROM teacher t1
WHERE t1.tname LIKE ‘李%‘
--5、查询没学过‘陈奕迅‘老师课的同学的学号、姓名；
/* 一对多关系的就会出现比较坑爹的情况，记得用distinct并且要考虑是不是要做一个子查询 */
/* 当正面来可能比较困难的时候就要反方面来进行 */
/* EXISTS版 */
SELECT t5.s# AS ST_CODE,
t5.sname AS ST_NAME
FROM student t5
WHERE NOT EXISTS (
SELECT 1
FROM (
SELECT DISTINCT t1.s# AS ST_CODE,
t1.sname AS ST_NAME,
t4.tname
FROM student t1
LEFT JOIN sc t2
ON t1.s# = t1.s#
LEFT JOIN course t3
ON t2.c# = t3.c#
LEFT JOIN teacher t4
ON t3.t# = t4.t#
WHERE t4.tname = ‘陈奕迅‘
)
WHERE s# = t5.s#
);
/* 子查询版 */
SELECT t5.s# AS ST_CODE,
t5.sname AS ST_NAME
FROM student t5
WHERE t5. EXISTS (
SELECT 1
FROM (
SELECT DISTINCT t1.s# AS ST_CODE,
t1.sname AS ST_NAME,
t4.tname
FROM student t1
LEFT JOIN sc t2
ON t1.s# = t1.s#
LEFT JOIN course t3
ON t2.c# = t3.c#
LEFT JOIN teacher t4
ON t3.t# = t4.t#
WHERE t4.tname = ‘陈奕迅‘
)
WHERE s# = t5.s#
);
--6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
--不能在where条件下面写WHERE t2.c# = 111 and t2.c# = 112 因为是对应同一条记录的
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
LEFT JOIN sc t2
ON t1.s# = t2.s#
WHERE t2.c# = 111
INTERSECT --集合的交集
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
LEFT JOIN sc t2
ON t1.s# = t2.s#
WHERE t2.c# = 112
--7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；
SELECT DISTINCT t1.s# AS ST_CODE,
t1.sname AS ST_NAME,
FROM student t1
LEFT JOIN sc t2
ON t1.s# = t1.s#
LEFT JOIN course t3
ON t2.c# = t3.c#
LEFT JOIN teacher t4
ON t3.t# = t4.t#
WHERE t4.tname = ‘李冰冰‘
--8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；
/* 用单纯的join版 */
/* 用了子查询 */
SELECT a.s#
FROM (SELECT t1.s#,t1.score FROM sc t1 WHERE t1.c# = 111) a
INNER JOIN (SELECT t2.s#,t2.score FROM sc t2 WHERE t2.c# = 112) b
ON a.s# = b.s#
WHERE a.score < b.score
--9、查询所有课程成绩小于60分的同学的学号、姓名；
SELECT T1.S#, T1.SNAME
FROM STUDENT T1
INNER JOIN SC T2
ON T1.S# = T2.S#
WHERE T2.C# < 60
--10、查询没有学全所有课的同学的学号、姓名；
/*子查询版*/
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN SC t2
ON t1.s# = t2.s#
WHERE t2.c# IN (
SELECT t3.c#
FROM Course t3
)
/*exists版*/
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN SC t2
ON t1.s# = t2.s#
WHERE EXISTS (
SELECT 1
FROM Course t3
WHERE t3.c# = t2.c#
)
--11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名；
/* 子查询 */
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
WHERE t2.c# IN (
SELECT t4.c#
FROM student t3
INNER JOIN sc t4
ON t3.s# = t4.s#
WHERE t3.s# = 1
)
/* exists版 */
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
WHERE EXISTS (
SELECT 1
FROM student t3
INNER JOIN sc t4
ON t3.s# = t4.s#
WHERE t4.c# = t2.c#
AND t3.s# = 1
)
--12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；
/* 子查询 */
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
AND t1.s# <> 1
WHERE t2.c# IN (
SELECT t4.c#
FROM student t3
INNER JOIN sc t4
ON t3.s# = t4.s#
WHERE t3.s# = 1
)
/* exists版 */
SELECT DISTINCT t1.s#,t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
AND t1.s# <> 1
WHERE EXISTS (
SELECT 1
FROM student t3
INNER JOIN sc t4
ON t3.s# = t4.s#
WHERE t4.c# = t2.c#
AND t3.s# = 1
)
--13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩；
UPDATE sc t4
SET t4.score = (
WITH tmp AS
(
SELECT t1.c#,
AVG(t1.score) AS avg_score
FROM sc t1
INNER JOIN Course t2
ON t1.c# = t2.c#
INNER JOIN teacher t3
ON t2.t# = t3.t#
GROUP BY t1.c#
)
SELECT t5.avg_score
FROM tmp t5
WHERE t4.c# = t5.c#
)
WHERE EXISTS (
SELECT 1
FROM sc t6
INNER JOIN Course t7
ON t6.c# = t7.c#
INNER JOIN teacher t8
ON t7.t# = t8.t#
AND t8.tname = ‘陈奕迅‘
)
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名；
SELECT DISTINCT t1.s#,
t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
AND t2.c# = ALL (
SELECT t4.c#
FROM student t3
INNER JOIN sc t4
ON t3.s# = t4.s#
AND t4.s# = 7
)
AND t1.s# <> 7
--15、删除学习‘陈奕迅‘老师课的SC表记录；
DELETE FROM sc t1
WHERE EXISTS (
SELECT 1
FROM teacher t2
INNER JOIN Course t3
ON t2.t# = t3.t#
WHERE t3.c# = t1.c#
AND t2.tname = ‘陈奕迅‘
)
16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2号课的平均成绩；
INSERT INTO sc(s#,c#,score)
SELECT min(t5.s#),112,AVG(t4.score)
FROM sc t4
INNER JOIN student t5
ON t4.s# = t5.s#
WHERE t4.s# NOT IN (
SELECT t2.s#
FROM student t2
INNER JOIN sc t3
ON t2.s# = t3.s#
AND t3.c# = 113
)
AND t4.c# = 112
GROUP BY t4.c#
--17、按平均成绩从高到低显示所有学生的“语文1”、“生物2”、“化学1”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
/* 行转列除了可以用poivt外还可以分别select出来，再和主表的字段关联 */
SELECT t1.S# as 学生ID,
(SELECT score FROM SC t2 WHERE t2.S#=t1.S# AND t2.C#=111) AS 语文1,
(SELECT score FROM SC t3 WHERE t3.S#=t1.S# AND t3.C#=118) AS 生物2,
(SELECT score FROM SC t4 WHERE t4.S#=t1.S# AND t4.C#=121) AS 化学1,
COUNT(*) AS 有效课程数,
AVG(t1.score) AS 平均成绩
FROM SC t1
GROUP BY t1.S#
ORDER BY avg(t1.score)
--18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分
SELECT t1.c#,
MIN(t1.score) AS min_score,
MAX(t1.score) AS max_score
FROM sc t1
GROUP BY t1.c#
--19、按各科平均成绩从低到高和及格率的百分数从高到低顺序
SELECT t1.c#,
AVG(t1.score) AS avg_score,
100 * SUM(
CASE
WHEN nvl(t1.score,0)>=60 THEN
1
ELSE
0
END
)/COUNT(*) AS pass_pct
FROM sc t1
GROUP BY t1.c#
ORDER BY avg_score ASC,pass_pct DESC
--20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 语文1（111），语文2（112），数学1（113），数学2（114）
SELECT SUM(CASE WHEN t1.c# =111 THEN score ELSE 0 END)/SUM(CASE t1.c# WHEN 111 THEN 1 ELSE 0 END) AS 语文1 ,
100 * SUM(CASE WHEN t1.c# = 111 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN t1.c# = 111 THEN 1 ELSE 0 END) AS 语文1及格百分比 ,
SUM(CASE WHEN t1.c# = 112 THEN score ELSE 0 END)/SUM(CASE t1.c# WHEN 112 THEN 1 ELSE 0 END) AS 语文2 ,
100 * SUM(CASE WHEN t1.c# = 112 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN t1.c# = 112 THEN 1 ELSE 0 END) AS 语文2及格百分比 ,
SUM(CASE WHEN t1.c# = 113 THEN score ELSE 0 END)/SUM(CASE t1.c# WHEN 113 THEN 1 ELSE 0 END) AS 数学1 ,
100 * SUM(CASE WHEN t1.c# = 113 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN t1.c# = 113 THEN 1 ELSE 0 END) AS 数学1及格百分比 ,
SUM(CASE WHEN t1.c# = 114 THEN score ELSE 0 END)/SUM(CASE t1.c# WHEN 114 THEN 1 ELSE 0 END) AS 数学2 ,
100 * SUM(CASE WHEN t1.c# = 114 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN t1.c# = 114 THEN 1 ELSE 0 END) AS 数学2及格百分比
FROM SC t1
--21、查询不同老师所教不同课程平均分从高到低显示
SELECT t3.tname,
t2.cname,
AVG(t1.score) AS avg_score
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
INNER JOIN teacher t3
ON t2.t# = t3.t#
GROUP BY t3.tname,t2.cname
ORDER BY avg_score DESC
--22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（111），马克思（112），UML （113），数据库（114）
SELECT sub_query.cname,
sub_query.rank
FROM(
SELECT t2.cname,
t1.score,
rank() over(PARTITION BY t1.c# ORDER BY t1.score DESC) AS rank
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
WHERE t1.c# IN (111,112,113,114)
) sub_query
WHERE sub_query.rank BETWEEN 3 AND 6;
--23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85] 优秀人数,[85-70] 良好人数,[70-60] 一般人数,[ <60] 刚及格人数
SELECT t1.c#,
t2.cname,
SUM(CASE WHEN t1.score > 85 AND t1.score <= 100 THEN 1 ELSE 0 END) AS 优秀人数,
SUM(CASE WHEN t1.score > 70 AND t1.score <= 85 THEN 1 ELSE 0 END) AS 良好人数,
SUM(CASE WHEN t1.score > 60 AND t1.score <= 70 THEN 1 ELSE 0 END) AS 一般人数,
SUM(CASE WHEN t1.score <= 60 THEN 1 ELSE 0 END) AS 刚及格人数
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
GROUP BY t1.c#,t2.cname
--24、查询学生平均成绩及其名次
SELECT t1.sname,
t1.s#,
AVG(t2.score) AS avg_score,
dense_rank() over(ORDER BY AVG(t2.score) DESC) AS dense_rank
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
GROUP BY t1.sname,t1.s#
--25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT sub_query.cname,
sub_query.s#,
sub_query.sname,
sub_query.score,
sub_query.rank
FROM (
SELECT t3.cname,
t1.s#,
t1.sname,
t2.score,
rank() over(PARTITION BY t2.c# ORDER BY t2.score DESC) AS rank
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
) sub_query
WHERE sub_query.rank < 4
--26、查询每门课程被选修的学生数
SELECT t2.cname,
COUNT(t1.s#) AS sum_stu
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
GROUP BY t2.cname
--27、查询出只选修了一门课程的全部学生的学号和姓名
/* GROUP BY 语句中选择出来的只有备份组的字段和聚合函数 */
SELECT t1.s#,
t1.sname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
GROUP BY t1.s#,t1.sname
HAVING COUNT(t2.c#) = 1
--28、查询男生、女生人数
SELECT t1.ssex,
COUNT(t1.ssex) AS sex_num
FROM student t1
GROUP BY t1.ssex
--29、查询名字中有‘黑‘的学生名单
SELECT t1.s#,
t1.sname
FROM student t1
WHERE t1.sname LIKE ‘%黑%‘
--30、查询同名同性学生名单，并统计同名人数
/* 写了个错的
原因：student表中有两个小粉，on条件只是县限定了sname相等，所以导致了2X2的笛卡尔乘积*/
SELECT t1.sname,
COUNT(t1.s#) AS same_name_count
FROM student t1
INNER JOIN student t2
ON t1.sname = t2.sname
GROUP BY t1.sname,t2.sname
HAVING COUNT(t1.sname) > 1
/* 正确的写法 */
SELECT t1.sname,
COUNT(t1.sname) AS same_name_count
FROM student t1
GROUP BY t1.sname
HAVING COUNT(t1.sname) > 1
--32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
SELECT t2.cname,
t2.c#,
AVG(t1.score) AS avg_score
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
GROUP BY t2.cname,t2.c#
ORDER BY avg_score ASC,t2.c# DESC
--33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
SELECT t1.s#,
t1.sname,
AVG(t2.score) AS avg_score
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
GROUP BY t1.s#,t1.sname
HAVING AVG(t2.score) > 85
--34、查询课程名称为‘语文1‘，且分数低于60的学生姓名和分数
SELECT t1.sname,
t2.score
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
WHERE t3.cname = ‘语文1‘
AND t2.score < 60
--35、查询所有学生的选课情况；
/* 可能不是这样写，有更好的写法 */
SELECT t1.s#,
t1.sname,
t3.cname
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
--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 t1.s#,
t1.sname,
t3.cname,
t2.score
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
WHERE t2.score > 70
--37、查询不及格的课程，并按课程号从大到小排列
SELECT t1.c#,
t2.cname,
t1.score
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
WHERE t1.score < 60
ORDER BY t1.c# DESC
--38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；
SELECT t1.s#,
t1.sname,
t2.score
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
WHERE t2.c# = 113
AND t2.score > 80
--39、求选了课程的学生人数
SELECT SUM(1)
FROM (
SELECT 1
FROM sc t1
GROUP BY t1.s#
)
--40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩
SELECT t4.s#,
t4.sname,
t5.score,
t6.cname
FROM student t4
INNER JOIN sc t5
ON t4.s# = t5.s#
INNER JOIN course t6
ON t5.c# = t6.c#
WHERE (t5.score,t6.cname) IN (
SELECT MAX(t1.score),
t2.cname
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
INNER JOIN teacher t3
ON t2.t# = t3.t#
WHERE t3.tname = ‘陈奕迅‘
GROUP BY t2.cname
)
--41、查询各个课程及相应的选修人数
SELECT t1.c#,
t2.cname,
COUNT(t1.s#)
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
GROUP BY t1.c#,t2.cname
--42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
SELECT t1.s#,
t1.c#,
t1.score
FROM sc t1
INNER JOIN sc t2
ON t1.score = t2.score
AND t1.c# <> t2.c#
AND t1.s# = t2.s#
--43、查询每门功成绩最好的前两名
SELECT sub_query.s#,
sub_query.sname,
sub_query.cname,
sub_query.score,
sub_query.rank
FROM (
SELECT t1.s#,
t3.sname,
t2.cname,
t1.score,
dense_rank() over(PARTITION BY t1.c# ORDER BY t1.score DESC) AS rank
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
INNER JOIN student t3
ON t1.s# = t3.s#
) sub_query
WHERE sub_query.rank < 3
--44、统计每门课程的学生选修人数（超过10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，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 t1.c#,
COUNT(t1.s#) AS stu_num
FROM sc t1
GROUP BY t1.c#
HAVING COUNT(t1.s#) > 10
ORDER BY stu_num DESC,t1.c# ASC
--45、检索至少选修两门课程的学生学号
SELECT t1.s#
FROM sc t1
GROUP BY t1.s#
HAVING COUNT(t1.c#) > 1
--46、查询全部学生都选修的课程的课程号和课程名
SELECT t1.c#,
t2.cname,
COUNT(t1.s#)
FROM sc t1
INNER JOIN course t2
ON t1.c# = t2.c#
GROUP BY t1.c#,t2.cname
HAVING COUNT(t1.s#) = (
SELECT COUNT(t3.s#)
FROM student t3
)
--47、查询没学过‘陈奕迅‘老师讲授的任一门课程的学生姓名
SELECT t5.s#,
t5.sname
FROM student t5
WHERE t5.s# NOT IN (
SELECT DISTINCT t1.s#
FROM student t1
INNER JOIN sc t2
ON t1.s# = t2.s#
INNER JOIN course t3
ON t2.c# = t3.c#
INNER JOIN teacher t4
ON t3.t# = t4.t#
WHERE t4.tname = ‘陈奕迅‘
)
--48、查询两门以上不及格课程的同学的学号及其平均成绩
/* 不能直接用下面的方式，因为有where条件过滤了一部分的成绩
SELECT t1.s#,
AVG(t1.score)
FROM sc t1
WHERE t1.score < 60
GROUP BY t1.s#
HAVING COUNT(t1.c#) > 2 */
SELECT t2.s#,
AVG(t2.score)
FROM sc t2
WHERE t2.s# IN (
SELECT t1.s#
FROM sc t1
WHERE t1.score < 60
GROUP BY t1.s#
HAVING COUNT(t1.c#) > 2
)
GROUP BY t2.s#
--49、检索‘114‘课程分数小于60，按分数降序排列的同学学号
SELECT t1.s#
FROM sc t1
WHERE t1.c# = 114
AND t1.score < 60
ORDER BY t1.s# DESC
--50、删除‘2‘同学的‘111‘课程的成绩
DELETE FROM sc t1
WHERE t1.s# = 2
AND t1.c# = 111
sql题
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