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