SQL面试题

饭宝 · 2025-2-12 18:29:59

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门生表

Student(S#,Sname,Sage,Ssex) 门生表

Course(C#,Cname,T#) 课程表

SC(S#,C#,score) 成绩表

Teacher(T#,Tname) 教师表

-- 1、查询“001”课程比“002”课程成绩高的所有学生的学号
SELECT a.S#
FROM (SELECT S#, score FROM SC WHERE C# = '001') a,
(SELECT S#, score FROM SC WHERE C# = '002') b
WHERE a.score > b.score AND a.S# = b.S#;
-- 2、查询平均成绩大于60分的同学的学号和平均成绩
SELECT S#, AVG(score) AS avg_score
FROM SC
GROUP BY S#
HAVING AVG(score) > 60;
-- 3、查询所有同学的学号、姓名、选课数、总成绩
SELECT Student.S#, Student.Sname, COUNT(SC.C#) AS course_count, SUM(score) AS total_score
FROM Student
LEFT JOIN SC ON Student.S# = SC.S#
GROUP BY Student.S#, Student.Sname;
-- 4、查询姓“李”的老师的个数
SELECT COUNT(DISTINCT Tname) AS teacher_count
FROM Teacher
WHERE Tname LIKE '李%';
-- 5、查询没学过“叶平”老师课的同学的学号、姓名
SELECT Student.S#, Student.Sname
FROM Student
WHERE S# NOT IN (
SELECT DISTINCT SC.S#
FROM SC, Course, Teacher
WHERE SC.C# = Course.C#
AND Teacher.T# = Course.T#
AND Teacher.Tname = '叶平'
);
-- 6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名
SELECT Student.S#, Student.Sname
FROM Student, SC
WHERE Student.S# = SC.S#
AND SC.C# = '001'
AND EXISTS (
SELECT 1
FROM SC AS SC_2
WHERE SC_2.S# = SC.S#
AND SC_2.C# = '002'
);
-- 7、查询学过“叶平”老师所教的所有课的同学的学号、姓名
SELECT S#, Sname
FROM Student
WHERE S# IN (
SELECT S#
FROM SC, Course, Teacher
WHERE SC.C# = Course.C#
AND Teacher.T# = Course.T#
AND Teacher.Tname = '叶平'
GROUP BY S#
HAVING COUNT(SC.C#) = (
SELECT COUNT(C#)
FROM Course, Teacher
WHERE Teacher.T# = Course.T#
AND Tname = '叶平'
)
);
-- 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
SELECT S#, Sname
FROM (
SELECT Student.S#, Student.Sname, score,
(SELECT score FROM SC SC_2 WHERE SC_2.S# = Student.S# AND SC_2.C# = '002') AS score2
FROM Student, SC
WHERE Student.S# = SC.S# AND C# = '001'
) AS S_2
WHERE score2 < score;
-- 9、查询所有课程成绩小于60分的同学的学号、姓名
SELECT S#, Sname
FROM Student
WHERE S# NOT IN (
SELECT S#
FROM SC
WHERE score > 60
);
-- 10、查询没有学全所有课的同学的学号、姓名
SELECT Student.S#, Student.Sname
FROM Student, SC
WHERE Student.S# = SC.S#
GROUP BY Student.S#, Student.Sname
HAVING COUNT(C#) < (SELECT COUNT(C#) FROM Course);
-- 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
SELECT S#, Sname
FROM Student, SC
WHERE Student.S# = SC.S#
AND C# IN (
SELECT C#
FROM SC
WHERE S# = '1001'
);
-- 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
SELECT DISTINCT SC.S#, Student.Sname
FROM Student, SC
WHERE Student.S# = SC.S#
AND SC.C# IN (SELECT C# FROM SC WHERE S# = '001')
AND Student.S# != '001';
-- 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
SET score = (
SELECT AVG(SC_2.score)
FROM SC SC_2
WHERE SC_2.C# = SC.C#
)
FROM SC, Course, Teacher
WHERE SC.C# = Course.C#
AND Course.T# = Teacher.T#
AND Teacher.Tname = '叶平';
-- 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 S#
FROM SC
WHERE C# IN (SELECT C# FROM SC WHERE S# = '1002')
GROUP BY S#
HAVING COUNT(*) = (SELECT COUNT(*) FROM SC WHERE S# = '1002')
AND S# != '1002';
-- 15、删除学习“叶平”老师课的SC表记录
DELETE FROM SC
FROM SC, Course, Teacher
WHERE SC.C# = Course.C#
AND Course.T# = Teacher.T#
AND Teacher.Tname = '叶平';
-- 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 S#, '002', (SELECT AVG(score) FROM SC WHERE C# = '002')
FROM Student
WHERE S# NOT IN (SELECT S# FROM SC WHERE C# = '003');
-- 17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩
SELECT
S# AS 学生ID,
(SELECT score FROM SC WHERE SC.S# = t.S# AND C# = '004') AS 数据库,
(SELECT score FROM SC WHERE SC.S# = t.S# AND C# = '001') AS 企业管理,
(SELECT score FROM SC WHERE SC.S# = t.S# AND C# = '006') AS 英语,
COUNT(*) AS 有效课程数,
AVG(t.score) AS 平均成绩
FROM SC AS t
GROUP BY S#
ORDER BY AVG(t.score) DESC;
-- 18、查询各科成绩最高和最低的分
SELECT
L.C# AS 课程ID,
MAX(L.score) AS 最高分,
MIN(R.score) AS 最低分
FROM SC L
JOIN SC R ON L.C# = R.C#
GROUP BY L.C#
HAVING MAX(L.score) = (SELECT MAX(IL.score) FROM SC IL WHERE IL.C# = L.C#)
AND MIN(R.score) = (SELECT MIN(IR.score) FROM SC IR WHERE IR.C# = R.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
t.C# AS 课程号,
MAX(course.Cname) AS 课程名,
ISNULL(AVG(score), 0) AS 平均成绩,
100 * SUM(CASE WHEN ISNULL(score, 0) >= 60 THEN 1 ELSE 0 END) / COUNT(*) AS 及格百分数
FROM SC t
JOIN Course ON t.C# = course.C#
GROUP BY t.C#
ORDER BY AVG(score) ASC, 100 * SUM(CASE WHEN ISNULL(score, 0) >= 60 THEN 1 ELSE 0 END) / COUNT(*) 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): 企业管理(001)，马克思(002)，OO&UML (003)，数据库(004)
SELECT
AVG(CASE WHEN C# = '001' THEN score ELSE NULL END) AS 企业管理平均分,
100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END) / SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数,
AVG(CASE WHEN C# = '002' THEN score ELSE NULL END) AS 马克思平均分,
100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END) / SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数,
AVG(CASE WHEN C# = '003' THEN score ELSE NULL END) AS UML平均分,
100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END) / SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数,
AVG(CASE WHEN C# = '004' THEN score ELSE NULL END) AS 数据库平均分,
100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END) / SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数
FROM SC;
-- 21、查询不同老师所教不同课程平均分从高到低显示
SELECT
MAX(Z.T#) AS 教师ID,
MAX(Z.Tname) AS 教师姓名,
C.C# AS 课程ID,
MAX(C.Cname) AS 课程名称,
AVG(T.score) AS 平均成绩
FROM SC AS T
JOIN Course AS C ON T.C# = C.C#
JOIN Teacher AS Z ON C.T# = Z.T#
GROUP BY C.C#
ORDER BY AVG(T.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(001)，马克思(002)，UML (003)，数据库(004)
-- [学生ID], [学生姓名], 企业管理, 马克思, UML, 数据库, 平均成绩
WITH StudentScores AS (
SELECT
SC.S# AS 学生学号,
Student.Sname AS 学生姓名,
SUM(CASE WHEN SC.C# = '001' THEN score ELSE 0 END) AS 企业管理,
SUM(CASE WHEN SC.C# = '002' THEN score ELSE 0 END) AS 马克思,
SUM(CASE WHEN SC.C# = '003' THEN score ELSE 0 END) AS UML,
SUM(CASE WHEN SC.C# = '004' THEN score ELSE 0 END) AS 数据库,
AVG(score) AS 平均成绩,
ROW_NUMBER() OVER (ORDER BY AVG(score) DESC) AS 排名
FROM Student
JOIN SC ON Student.S# = SC.S#
GROUP BY SC.S#, Student.Sname
)
SELECT
学生学号,
学生姓名,
企业管理,
马克思,
UML,
数据库,
平均成绩
FROM StudentScores
WHERE 排名 BETWEEN 3 AND 6;
-- 29、查询姓“张”的学生名单
SELECT Sname
FROM Student
WHERE Sname LIKE '张%';
-- 30、查询同名同性学生名单，并统计同名人数
SELECT Sname, COUNT(*)
FROM Student
GROUP BY Sname
HAVING COUNT(*) > 1;
-- 31、1981年出生的学生名单 (注: Student 表中 Sage 列的类型是 datetime)
SELECT Sname, CONVERT(CHAR(4), DATEPART(YEAR, Sage)) AS BirthYear
FROM Student
WHERE CONVERT(CHAR(4), DATEPART(YEAR, Sage)) = '1981';
-- 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 C#, AVG(score) AS AvgScore
FROM SC
GROUP BY C#
ORDER BY AvgScore, 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 Student.Sname, SC.S#, AVG(SC.score) AS AvgScore
FROM Student
JOIN SC ON Student.S# = SC.S#
GROUP BY SC.S#, Student.Sname
HAVING AVG(SC.score) > 85;
-- 34、查询课程名称为“数据库”，且分数低于60的学生姓名和分数
SELECT Student.Sname, ISNULL(SC.score, 0) AS Score
FROM Student
JOIN SC ON Student.S# = SC.S#
JOIN Course ON SC.C# = Course.C#
WHERE Course.Cname = '数据库' AND SC.score < 60;
-- 35、查询所有学生的选课情况
SELECT SC.S#, SC.C#, Student.Sname, Course.Cname
FROM SC
JOIN Student ON SC.S# = Student.S#
JOIN Course ON SC.C# = Course.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 DISTINCT Student.Sname, Course.Cname, SC.score
FROM Student
JOIN SC ON Student.S# = SC.S#
JOIN Course ON SC.C# = Course.C#
WHERE SC.score >= 70;
-- 37、查询不及格的课程，并按课程号从大到小排列
SELECT C#
FROM SC
WHERE score < 60
ORDER BY 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 SC.S#, Student.Sname
FROM SC
JOIN Student ON SC.S# = Student.S#
WHERE SC.score > 80 AND SC.C# = '003';
-- 39、求选了课程的学生人数
SELECT COUNT(DISTINCT S#)
FROM SC;
-- 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 Student.Sname, SC.score
FROM Student
JOIN SC ON Student.S# = SC.S#
JOIN Course ON SC.C# = Course.C#
JOIN Teacher ON Course.T# = Teacher.T#
WHERE Teacher.Tname = '叶平'
AND SC.score = (SELECT MAX(score) FROM SC SC_INNER WHERE SC_INNER.C# = Course.C#);
-- 41、查询各个课程及相应的选修人数
SELECT C#, COUNT(*) AS StudentCount
FROM SC
GROUP BY C#;
-- 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 DISTINCT A.S#, A.C# AS CourseA, B.C# AS CourseB, A.score
FROM SC A
JOIN SC B ON A.score = B.score AND A.C# <> B.C#;
-- 43、查询每门课程成绩最好的前两名
SELECT t1.S# AS 学生ID, t1.C# AS 课程ID, t1.score AS 分数
FROM SC t1
WHERE t1.score IN (
SELECT TOP 2 score
FROM SC
WHERE C# = t1.C#
ORDER BY score DESC
)
ORDER BY t1.C#, t1.score DESC;
-- 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，若人数相同，按课程号升序排列
SELECT
C# AS 课程号,
COUNT(*) AS 人数
FROM
SC
GROUP BY
C#
HAVING
COUNT(*) > 10
ORDER BY
人数 DESC,
课程号;
-- 45、检索至少选修两门课程的学生学号
SELECT
S#
FROM
SC
GROUP BY
S#
HAVING
COUNT(*) >= 2;
-- 46、查询全部学生都选修的课程的课程号和课程名
SELECT
C#,
Cname
FROM
Course
WHERE
C# IN (
SELECT
C#
FROM
SC
GROUP BY
C#
HAVING
COUNT(DISTINCT S#) = (SELECT COUNT(*) FROM Student)
);
-- 47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
SELECT
Sname
FROM
Student
WHERE
S# NOT IN (
SELECT
SC.S#
FROM
Course,
Teacher,
SC
WHERE
Course.T# = Teacher.T#
AND SC.C# = Course.C#
AND Teacher.Tname = '叶平'
);
-- 48、查询两门以上不及格课程的同学的学号及其平均成绩
SELECT
S#,
AVG(ISNULL(score, 0)) AS 平均成绩
FROM
SC
WHERE
S# IN (
SELECT
S#
FROM
SC
WHERE
score < 60
GROUP BY
S#
HAVING
COUNT(*) > 2
)
GROUP BY
S#;
-- 49、检索“004”课程分数小于60，按分数降序排列的同学学号
SELECT
S#
FROM
SC
WHERE
C# = '004'
AND score < 60
ORDER BY
score DESC;
-- 50、删除“002”同学的“001”课程的成绩
DELETE FROM
SC
WHERE
S# = '002'
AND C# = '001';

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图书表

本题用到下面三个关系表:

CARD 借书卡。 CNO 卡号，NAME 姓名，CLASS 班级
BOOKS 图书。 BNO 书号，BNAME 书名,AUTHOR 作者，PRICE 单价，QUANTITY 库 存册数
BORROW 借书记载。 CNO 借书卡号，BNO 书号，RDATE 还书日期备注:限定每人每种书只能借一本;库存册数随借书、还书而改变。

-- 1. 建立 BORROW 表的 SQL 语句，定义主码完整性约束和引用完整性约束
CREATE TABLE BORROW (
CNO int,
BNO int,
RDATE datetime,
FOREIGN KEY (CNO) REFERENCES CARD(CNO),
FOREIGN KEY (BNO) REFERENCES BOOKS(BNO),
PRIMARY KEY (CNO, BNO)
);
-- 2. 找出借书超过5本的读者，输出借书卡号及所借图书册数
SELECT
CNO,
COUNT(*) AS 借图书册数
FROM
BORROW
GROUP BY
CNO
HAVING
COUNT(*) > 5;
-- 3. 查询借阅了"水浒"一书的读者，输出姓名及班级
SELECT
c.NAME,
c.CLASS
FROM
CARD c
WHERE
EXISTS (
SELECT 1
FROM
BORROW a
JOIN
BOOKS b ON a.BNO = b.BNO
WHERE
b.BNAME = N'水浒'
AND a.CNO = c.CNO
);
-- 4. 查询过期未还图书，输出借阅者(卡号)、书号及还书日期
SELECT
CNO AS 借阅者卡号,
BNO AS 书号,
RDATE AS 还书日期
FROM
BORROW
WHERE
RDATE < GETDATE();
-- 5. 查询书名包括"网络"关键词的图书，输出书号、书名、作者
SELECT
BNO,
BNAME,
AUTHOR
FROM
BOOKS
WHERE
BNAME LIKE N'%网络%';
-- 6. 查询现有图书中价格最高的图书，输出书名及作者
SELECT
BNAME,
AUTHOR
FROM
BOOKS
WHERE
PRICE = (SELECT MAX(PRICE) FROM BOOKS);
-- 7. 查询当前借了"计算方法"但没有借"计算方法习题集"的读者，输出其借书卡号，并按卡号降序排序
SELECT
a.CNO
FROM
BORROW a
JOIN
BOOKS b ON a.BNO = b.BNO
WHERE
b.BNAME = N'计算方法'
AND NOT EXISTS (
SELECT 1
FROM
BORROW aa
JOIN
BOOKS bb ON aa.BNO = bb.BNO
WHERE
bb.BNAME = N'计算方法习题集'
AND aa.CNO = a.CNO
)
ORDER BY
a.CNO DESC;
-- 8. 将"C01"班同学所借图书的还期都延长一周
UPDATE b
SET
RDATE = DATEADD(Day, 7, b.RDATE)
FROM
CARD a
JOIN
BORROW b ON a.CNO = b.CNO
WHERE
a.CLASS = N'C01';
-- 9. 从 BOOKS 表中删除当前无人借阅的图书记录
DELETE FROM BOOKS
WHERE
NOT EXISTS (
SELECT 1
FROM BORROW
WHERE BNO = BOOKS.BNO
);
-- 10. 如果经常按书名查询图书信息，建立合适的索引
CREATE CLUSTERED INDEX IDX_BOOKS_BNAME ON BOOKS (BNAME);
-- 11. 在 BORROW 表上建立触发器，保存借阅"数据库技术及应用"的读者记录到 BORROW_SAVE 表
CREATE TRIGGER TR_SAVE ON BORROW
AFTER INSERT, UPDATE
AS
BEGIN
IF @@ROWCOUNT > 0
BEGIN
INSERT INTO BORROW_SAVE
SELECT
i.*
FROM
INSERTED i
JOIN
BOOKS b ON i.BNO = b.BNO
WHERE
b.BNAME = N'数据库技术及应用';
END
END;
-- 12. 建立视图，显示"力01"班学生的借书信息(姓名和书名)
CREATE VIEW V_VIEW AS
SELECT
a.NAME,
b.BNAME AS 书名
FROM
BORROW ab
JOIN
CARD a ON ab.CNO = a.CNO
JOIN
BOOKS b ON ab.BNO = b.BNO
WHERE
a.CLASS = N'力01';
-- 13. 查询当前同时借有"计算方法"和"组合数学"的读者，输出其借书卡号，并按卡号升序排序
SELECT
a.CNO
FROM
BORROW a
JOIN
BOOKS b ON a.BNO = b.BNO
WHERE
b.BNAME IN (N'计算方法', N'组合数学')
GROUP BY
a.CNO
HAVING
COUNT(DISTINCT b.BNAME) = 2
ORDER BY
a.CNO;
-- 14. 假定在建 BOOKS 表时没有定义主码，写出为 BOOKS 表追加定义主码的语句
ALTER TABLE BOOKS
ADD PRIMARY KEY (BNO);
-- 15.1 将 NAME 最大列宽增加到10个字符(假定原为6个字符)
ALTER TABLE CARD
ALTER COLUMN NAME varchar(10);
-- 15.2 为该表增加1列 NAME(系名)，可变长，最大20个字符
ALTER TABLE CARD
ADD 系名 varchar(20);

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管理表

为管理岗位业务培训信息，创建3个表:

S (S#,SN,SD,SA)
C (C#,CN )
SC ( S#,C#,G )
S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年事 C#,CN 分别代表课程编号、课程名称
S#,C#,G 分别代表学号、所选修的课程编号、学习成绩

-- 1. 查询选修课程名称为’税收基础’的学员学号和姓名
SELECT
S.SN,
S.SD
FROM
S
WHERE
S.[S#] IN (
SELECT
SC.[S#]
FROM
C,
SC
WHERE
C.[C#] = SC.[C#]
AND C.CN = N'税收基础'
);
-- 2. 查询选修课程编号为’C2’的学员姓名和所属单位
SELECT
S.SN,
S.SD
FROM
S,
SC
WHERE
S.[S#] = SC.[S#]
AND SC.[C#] = 'C2';
-- 3. 查询不选修课程编号为’C5’的学员姓名和所属单位
SELECT
SN,
SD
FROM
S
WHERE
[S#] NOT IN (
SELECT
[S#]
FROM
SC
WHERE
[C#] = 'C5'
);
-- 4. 查询选修全部课程的学员姓名和所属单位
-- 注意：此查询逻辑可能需要根据实际数据库结构进行调整，因为RIGHT JOIN和COUNT(DISTINCT [S#])的使用可能不是最优解
SELECT
SN,
SD
FROM
S
WHERE
[S#] IN (
SELECT
[S#]
FROM
SC
RIGHT JOIN
C ON SC.[C#] = C.[C#]
GROUP BY
[S#]
HAVING
COUNT(*) = (SELECT COUNT(*) FROM C) -- 这里应该确保统计的是所有课程的数量
);
-- 5. 查询选修了课程的学员人数
SELECT
COUNT(DISTINCT [S#]) AS 学员人数
FROM
SC;
-- 6. 查询选修课程超过5门的学员学号和所属单位
SELECT
SN,
SD
FROM
S
WHERE
[S#] IN (
SELECT
[S#]
FROM
SC
GROUP BY
[S#]
HAVING
COUNT(DISTINCT [C#]) > 5
);
-- 以下是关于cj表的创建、数据插入和查询语句
-- 如果cj表已存在，则删除它
IF OBJECT_ID('cj') IS NOT NULL
DROP TABLE cj;
-- 创建cj表
CREATE TABLE cj (
stuName nvarchar(10),
KCM nvarchar(10),
cj numeric(5,2)
);
-- 向cj表中插入数据
INSERT INTO cj
SELECT '张三', '语文', 98
UNION
SELECT '李四', '语文', 89
UNION
SELECT '王五', '语文', 67
UNION
SELECT '周攻', '语文', 56
UNION
SELECT '张三', '数学', 89
UNION
SELECT '李四', '数学', 78
UNION
SELECT '王五', '数学', 90
UNION
SELECT '周攻', '数学', 87;
-- 方法一: 查询每门课程中成绩不是最低的学员姓名
SELECT
stuname
FROM (
SELECT
stuName,
kcm,
(SELECT COUNT(*) FROM cj WHERE stuname != a.stuname AND kcm = a.kcm AND cj > a.cj) cnt
FROM
cj a
) x
GROUP BY
stuname
HAVING
MAX(cnt) <= 1;
-- 方法二: 查询每门课程中成绩排名前二的学员姓名（如果有重复成绩，则可能返回多于两名）
SELECT
stuname
FROM
cj a
WHERE
cj IN (
SELECT TOP 2 cj
FROM
cj
WHERE
kcm = a.kcm
ORDER BY
cj DESC
)
GROUP BY
stuname
HAVING
(COUNT(1) > 1);
-- 方法三: 查询每门课程中成绩不是最差的学员姓名（另一种实现方式）
SELECT DISTINCT
stuname
FROM
cj a
WHERE
NOT EXISTS (
SELECT
kcm
FROM
cj b
WHERE
a.stuname = b.stuname
AND (SELECT COUNT(*) FROM cj WHERE kcm = b.kcm AND stuname != a.stuname AND cj > b.cj) > 1
);

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