首页笔试题SQL

从网上收集整理的 SQL 笔试题。

DBMS 无特殊说明的情况下使用 PostgreSQL

# 一

# 表结构预览

-- 学生表
Student (SId,Sname,Sage,Ssex)
--SId 学生编号，Sname 学生姓名，Sage 出生年月，Ssex 学生性别
-- 课程表
Course (CId,Cname,TId)
--CId 课程编号，Cname 课程名称，TId 教师编号
-- 教师表
Teacher (TId,Tname)
--TId 教师编号，Tname 教师姓名
-- 成绩表
SC (SId,CId,score)
--SId 学生编号，CId 课程编号，score 分数

# 题目

  1. 查询 “01” 课程比 “02” 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 “张三” 老师课的同学的学号、姓名
  6. 查询学过 “01” 并且也学过编号 “02” 课程的同学的学号、姓名
  7. 查询学过 “张三” 老师所教的课的同学的学号、姓名
  8. 查询课程编号 “01” 的成绩比课程编号 “02” 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 “01” 的同学所学相同的同学的学号和姓名
  12. 查询和 "01" 号的同学学习的课程完全相同的其他同学的学号和姓名
  13. 把 “SC” 表中 “张三” 老师教的课的成绩都更改为此课程的平均成绩
  14. 查询没学过 "张三" 老师讲授的任一门课程的学生姓名
  15. 查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
  16. 检索 "01" 课程分数小于 60，按分数降序排列的学生信息
  17. 按平均成绩从高到低显示所有学生的平均成绩
  18. 查询各科成绩最高分、最低分和平均分：以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率
  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. 查询学生的总成绩并进行排名
  21. 查询不同老师所教不同课程平均分从高到低显示
  22. 查询所有课程的成绩第 2 名到第 3 名的学生信息及该课程成绩
  23. 统计各科成绩各分数段人数：课程编号，课程名称，[100-85],[85-70],[70-60],[0-60] 及所占百分比
  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
  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. 查询名字中含有 "风" 字的学生信息
  30. 查询同名同性学生名单，并统计同名人数
  31. 查询 1990 年出生的学生名单 (注：Student 表中 Sage 列的类型是 datetime)
  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. 查询不及格的课程，并按课程号从大到小排列
  34. 查询课程编号为 "01" 且课程成绩在 60 分以上的学生的学号和姓名
  35. 查询选修 “张三” 老师所授课程的学生中，成绩最高的学生姓名及其成绩
  36. 查询每门功课成绩最好的前两名
  37. 统计每门课程的学生选修人数（超过 5 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
  38. 检索至少选修两门课程的学生学号
  39. 查询选修了全部课程的学生信息
  40. 查询各学生的年龄
  41. 查询本周过生日的学生
  42. 查询下周过生日的学生
  43. 查询本月过生日的学生
  44. 查询下月过生日的学生

-- 答案分割线 --

# 附录：答案

  1. 查询 “01” 课程比 “02” 课程成绩高的所有学生的学号
SELECT t1.sid AS 学号
FROM (SELECT * FROM sc WHERE CID = '01') AS t1
        LEFT JOIN (SELECT * FROM sc WHERE CID = '02') AS t2 ON t1.sid = t2.sid
WHERE t1.score > t2.score
  1. 查询平均成绩大于 60 分的同学的学号和平均成绩
SELECT sid        AS 学号,
      AVG(score) AS 平均成绩
FROM sc
GROUP BY sid
HAVING AVG(score) > 60
  1. 查询所有同学的学号、姓名、选课数、总成绩
SELECT S.sid      AS 学号,
      S.sname    AS 姓名,
      COUNT(CID) AS 选课数,
      SUM(score) AS 总成绩
FROM student AS S
        LEFT JOIN sc ON S.sid = SC.sid
GROUP BY S.sid,
        s.sname
  1. 查询姓 “李” 的老师的个数
SELECT COUNT(TID) AS teacher_cnt
FROM teacher
WHERE tname LIKE '李%'
  1. 查询没学过 “张三” 老师课的同学的学号、姓名
    使用 LEFT JOIN（原答案）：
SELECT sid,
      sname
FROM student
WHERE sid NOT IN (SELECT sc.sid
                  FROM teacher
                          LEFT JOIN course ON teacher.TID = course.TID
                          LEFT JOIN sc ON course.CID = sc.CID
                  WHERE teacher.tname = '张三')

或者使用 INNER JOIN：

SELECT sid   AS 学号,
      sname AS 姓名
FROM student
WHERE sid NOT IN (SELECT SC.sid
                  FROM SC,
                      teacher AS T,
                      course AS C
                  WHERE SC.CID = C.CID
                    AND T.TID = C.TID
                    AND T.tname = '张三')
  1. 查询学过 “01” 并且也学过编号 “02” 课程的同学的学号、姓名
SELECT T.sid AS 学号,
      sname AS 姓名
FROM (SELECT sid
      FROM sc
      GROUP BY sid
      HAVING SUM(CASE WHEN CID = '01' THEN 1 ELSE 0 END) > 0
        AND SUM(CASE WHEN CID = '02' THEN 1 ELSE 0 END) > 0) T
        LEFT JOIN student ON T.sid = student.sid
  1. 查询学过 “张三” 老师所教的课的同学的学号、姓名
SELECT S.sid   AS 学号,
      S.sname AS 姓名
FROM (SELECT cid
      FROM course AS C
              LEFT JOIN teacher AS T ON C.tid = T.tid
      WHERE T.tname = '张三') course
        LEFT JOIN sc ON course.cid = sc.cid
        LEFT JOIN student AS S ON sc.sid = S.sid
GROUP BY S.sid, sname
  1. 查询课程编号 “01” 的成绩比课程编号 “02” 课程低的所有同学的学号、姓名
SELECT S.sid   AS 学号,
      S.sname AS 姓名
FROM (SELECT t1.sid AS sid
      FROM (SELECT * FROM sc WHERE cid = '01') t1
              LEFT JOIN (SELECT * FROM sc WHERE cid = '02') t2 ON t1.sid = t2.sid
      WHERE t1.score < t2.score) t1
        LEFT JOIN student AS S ON t1.sid = S.sid
  1. 查询所有课程成绩小于 60 分的同学的学号、姓名
SELECT S.sid   AS 学号,
      S.sname AS 姓名
FROM (SELECT sid
      FROM sc
      GROUP BY sid
      HAVING MAX(score) < 60) t1
LEFT JOIN student AS S ON t1.sid = S.sid
  1. 查询没有学全所有课的同学的学号、姓名
SELECT S.sid   AS 学号,
      S.sname AS 姓名
FROM (SELECT COUNT(cid), sid
      FROM sc
      GROUP BY sid
      HAVING COUNT(cid) < (SELECT COUNT(cid) FROM course)) t1
LEFT JOIN student AS S ON t1.sid = S.sid
  1. 查询至少有一门课与学号为 “01” 的同学所学相同的同学的学号和姓名
SELECT S.sid   AS 学号,
      S.sname AS 姓名
FROM (SELECT cid
      FROM sc
      WHERE sid = '01') t1
LEFT JOIN sc ON t1.cid = sc.cid
LEFT JOIN student AS S ON S.sid = sc.sid
GROUP BY S.sid, s.sname
  1. 查询和 "01" 号的同学学习的课程完全相同的其他同学的学号和姓名
#注意是和 '01' 号同学课程完全相同但非学习课程数相同的
SELECT S.sid AS 学号, S.sname AS 姓名
FROM (SELECT sc.sid, COUNT(sc.cid)
      FROM (SELECT cid
            FROM sc
            WHERE sid = '01') t1 -- 选出 01 的同学所学的课程
      LEFT JOIN sc ON t1.cid = sc.cid
      GROUP BY sc.sid
      HAVING COUNT(sc.cid) = (SELECT COUNT(cid) FROM sc WHERE sid = '01')) t1
LEFT JOIN student AS S ON t1.sid = S.sid
WHERE S.sid != '01'
  1. 把 “SC” 表中 “张三” 老师教的课的成绩都更改为此课程的平均成绩
    这道题是更新数据的题目，不进行处理
  1. 查询没学过 "张三" 老师讲授的任一门课程的学生姓名
SELECT sname AS 姓名
FROM student
WHERE sid NOT IN
      (SELECT sid
      FROM sc
      LEFT JOIN course AS C ON sc.cid = C.cid
      LEFT JOIN teacher AS T ON C.tid = T.tid
      WHERE tname = '张三')
  1. 查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
SELECT S.sid   AS 学号,
      S.sname AS 姓名,
      平均成绩
FROM (SELECT sid,
            SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END),
            AVG(score) AS 平均成绩
      FROM sc
      GROUP BY sid
      HAVING SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) >= 2) t1
LEFT JOIN student AS S ON t1.sid = S.sid
  1. 检索 "01" 课程分数小于 60，按分数降序排列的学生信息
SELECT sid AS 学号, (CASE WHEN cid = '01' THEN score ELSE 100 END) AS 分数
FROM sc
WHERE (CASE WHEN cid = '01' THEN score ELSE 100 END) < 60
ORDER BY 分数 DESC
  1. 按平均成绩从高到低显示所有学生的平均成绩
SELECT sid AS 学号, AVG(score) AS 平均成绩
FROM sc
GROUP BY sid
ORDER BY 平均成绩 DESC
  1. 查询各科成绩最高分、最低分和平均分：以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率
SELECT c.cid                                                                                     AS 课程ID,
      c.cname                                                                                   AS 课程name,
      MAX(sc.score)                                                                             AS 最高分,
      MIN(sc.score)                                                                             AS 最低分,
      AVG(sc.score)                                                                             AS 平均分,
      SUM(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / CAST(COUNT(sc.sid) AS DOUBLE PRECISION) AS 及格率
FROM course AS C
LEFT JOIN sc ON c.cid = sc.cid
GROUP BY c.cid, c.cname
  1. 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
#这里先按照平均成绩排序，再按照及格百分数排序，
SELECT c.cid                                                                                     AS 课程ID,
      c.cname                                                                                   AS 课程name,
      AVG(sc.score)                                                                             AS 平均分,
      SUM(CASE WHEN sc.score >= 60 THEN 1 ELSE 0 END) / CAST(COUNT(sc.sid) AS DOUBLE PRECISION) AS 及格率
FROM course AS C
LEFT JOIN sc ON c.cid = sc.cid
GROUP BY c.cid, c.cname
ORDER BY 平均分, 及格率 DESC
  1. 查询学生的总成绩并进行排名
SELECT sid        AS 学号,
      SUM(score) AS 总成绩
FROM sc
GROUP BY sid
ORDER BY 总成绩 DESC
  1. 查询不同老师所教不同课程平均分从高到低显示
-- 每位老师只教一门课程，所以直接按老师 id 分组计算平均分即可
SELECT C.tid, AVG(sc.score) AS 平均成绩
FROM course AS C
  LEFT JOIN sc ON sc.cid = C.cid
GROUP BY C.tid
ORDER BY 平均成绩 DESC;
  1. 查询所有课程的成绩第 2 名到第 3 名的学生信息及该课程成绩
SELECT sid,
      rank_num,
      score,
      cid
FROM (SELECT RANK() OVER (PARTITION BY cid ORDER BY score DESC) AS rank_num,
            sid,
            score,
            cid
      FROM sc) t
WHERE rank_num IN (2, 3);
  1. 统计各科成绩各分数段人数：课程编号，课程名称，[100-85],[85-70],[70-60],[0-60] 及所占百分比
    （题目似乎有歧义，前边说统计各分数段人数，但是后边又说到百分比，个人认为是需要加上人数统计比较合理）
  • PostgreSQL：
SELECT sc.cid,
      cname,
      SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) / CAST(COUNT(sid) AS DOUBLE PRECISION) p1,
      SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) / CAST(COUNT(sid) AS DOUBLE PRECISION)  p2,
      SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) / CAST(COUNT(sid) AS DOUBLE PRECISION)  p3,
      SUM(CASE WHEN score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) / CAST(COUNT(sid) AS DOUBLE PRECISION)   p4
FROM sc
        LEFT JOIN course AS C ON sc.cid = C.cid
GROUP BY sc.cid, cname;
  • MySql：
select
  sc.cid
  ,cname
  ,count(if(score between 85 and 100,sid,null))/count(sid)
  ,count(if(score between 70 and 85,sid,null))/count(sid)
  ,count(if(score between 60 and 70,sid,null))/count(sid)
  ,count(if(score between 0 and 60,sid,null))/count(sid)
from sc
left join course
    on sc.cid=course.cid
group by sc.cid,cname
  1. 查询学生平均成绩及其名次
SELECT sid,
      avg_score,
      RANK() OVER (ORDER BY avg_score DESC ) AS rank
FROM (SELECT sid,
            AVG(score) avg_score
      FROM sc
      GROUP BY sid) t;
  1. 查询各科成绩前三名的记录
SELECT cid, sid, rank, score
FROM (SELECT cid,
            sid,
            RANK() OVER (PARTITION BY cid ORDER BY score DESC ) AS rank,
            score
      FROM sc) t
WHERE t.rank <= 3;
  1. 查询每门课程被选修的学生数
SELECT cid, COUNT(sid) AS cnt
FROM sc
GROUP BY cid;
  1. 查询出只选修了一门课程的全部学生的学号和姓名
SELECT S.sid, S.sname
FROM (SELECT sid
      FROM sc
      GROUP BY sid
      HAVING COUNT(cid) = 1) t1
LEFT JOIN student AS S ON S.sid = t1.sid;
  1. 查询男生、女生人数
SELECT ssex, COUNT(sid) AS cnt
FROM student
GROUP BY ssex;
  1. 查询名字中含有 "风" 字的学生信息
SELECT sid, sname
FROM student
WHERE sname LIKE '%风%';
  1. 查询同名同性学生名单，并统计同名人数
SELECT sname, ssex, COUNT(sid) AS cnt
FROM student
GROUP BY sname, ssex
HAVING COUNT(sid) >= 2;
  1. 查询 1990 年出生的学生名单 (注：Student 表中 Sage 列的类型是 datetime)
  • PostgreSQL：
SELECT sid, sname, sage
FROM student
WHERE DATE_PART('year', sage) = 1990;
  • MySql：
SELECT sid,
      sname,
      sage
FROM student
WHERE YEAR(sage) = 1990;
  1. 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 cid, AVG(score) AS avg_score
FROM sc
GROUP BY cid
ORDER BY avg_score, cid DESC;
  1. 查询不及格的课程，并按课程号从大到小排列
    (题目出得意义不是很明确)
SELECT cid,
      sid,
      score
FROM sc
WHERE score < 60
ORDER BY cid DESC, sid;
  1. 查询课程编号为 "01" 且课程成绩在 60 分以上的学生的学号和姓名
SELECT s.sid, s.sname
FROM sc
  LEFT JOIN student AS S ON S.sid = sc.sid
WHERE sc.cid = '01'
  AND sc.score >= 60;
  1. 查询选修 “张三” 老师所授课程的学生中，成绩最高的学生姓名及其成绩
SELECT sc.sid, sc.score
FROM teacher AS T
  LEFT JOIN course AS C ON T.tid = C.tid
  LEFT JOIN sc ON sc.cid = C.cid
  LEFT JOIN student AS S ON S.sid = sc.sid
WHERE T.tname = '张三'
ORDER BY sc.score DESC
LIMIT 1;
  1. 查询每门功课成绩最好的前两名
SELECT cid, sid, rank, score
FROM (SELECT cid,
            sid,
            RANK() OVER (PARTITION BY cid ORDER BY score DESC) AS rank,
            score
      FROM sc) t
WHERE t.rank <= 2;
  1. 统计每门课程的学生选修人数（超过 5 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 cid, COUNT(sid) AS cnt
FROM sc
GROUP BY cid
HAVING COUNT(sid) > 5
ORDER BY cnt DESC, cid;
  1. 检索至少选修两门课程的学生学号
SELECT sid, count(cid) as cnt
FROM sc
GROUP BY sid
HAVING COUNT(cid) >= 2;
  1. 查询选修了全部课程的学生信息
SELECT sid
FROM sc
GROUP BY sid
HAVING COUNT(cid) = (SELECT COUNT(*) FROM course);
  1. 查询各学生的年龄
  • PostgreSQL:
SELECT sid, sname, DATE_PART('year', CURRENT_DATE) - DATE_PART('year', sage) AS age
FROM student
  • MySql:
SELECT sid,
      sname,
      year(curdate()) - year(sage) AS sage
FROM student
  1. 查询本周过生日的学生
  • PostgreSQL:
SELECT sid, sname, sage
FROM student
WHERE DATE_PART('week', CURRENT_DATE) = DATE_PART('week', sage)
  • MySql：
select
  sid,sname,sage
from student
where weekofyear(sage)=weekofyear(curdate())
  1. 查询下周过生日的学生
  • PostgreSQL:
SELECT sid, sname, sage
FROM student
WHERE DATE_PART('week', CURRENT_DATE + INTERVAL '1 week') = DATE_PART('week', sage);
  • MySql:
select 
  sid,sname,sage
from student
where weekofyear(sage) = weekofyear(date_add(curdate(),interval 1 week))
  1. 查询本月过生日的学生
  • PostgreSQL:
SELECT sid, sname, sage
FROM student
WHERE DATE_PART('month', CURRENT_DATE) = DATE_PART('month', sage);
  • MySql:
select
  sid,sname,sage
from student
where month(sage) = month(curdate())
  1. 查询下月过生日的学生
  • PostgreSQL:
SELECT sid, sname, sage
FROM student
WHERE DATE_PART('month', CURRENT_DATE + INTERVAL '1 month') = DATE_PART('month', sage);
  • MySql:
select
  sid,sname,sage
from student
where month(date_sub(sage,interval 1 month)) = month(curdate())

# 附录：建表语句

  • PostgreSQL
CREATE TABLE Student ( SId VARCHAR ( 10 ), Sname VARCHAR ( 10 ), Sage TIMESTAMP, Ssex VARCHAR ( 10 ) );
INSERT INTO Student
VALUES
 ( '01', '赵雷', '1990-01-01', '男' );
INSERT INTO Student
VALUES
 ( '02', '钱电', '1990-12-21', '男' );
INSERT INTO Student
VALUES
 ( '03', '孙风', '1990-05-20', '男' );
INSERT INTO Student
VALUES
 ( '04', '李云', '1990-08-06', '男' );
INSERT INTO Student
VALUES
 ( '05', '周梅', '1991-12-01', '女' );
INSERT INTO Student
VALUES
 ( '06', '吴兰', '1992-03-01', '女' );
INSERT INTO Student
VALUES
 ( '07', '郑竹', '1989-07-01', '女' );
INSERT INTO Student
VALUES
 ( '08', '王菊', '1990-01-20', '女' );
CREATE TABLE Course ( CID VARCHAR ( 10 ), Cname VARCHAR ( 10 ), TID VARCHAR ( 10 ) );
INSERT INTO Course
VALUES
 ( '01', '语文', '02' );
INSERT INTO Course
VALUES
 ( '02', '数学', '01' );
INSERT INTO Course
VALUES
 ( '03', '英语', '03' );
CREATE TABLE Teacher ( TID VARCHAR ( 10 ), Tname VARCHAR ( 10 ) );
INSERT INTO Teacher
VALUES
 ( '01', '张三' );
INSERT INTO Teacher
VALUES
 ( '02', '李四' );
INSERT INTO Teacher
VALUES
 ( '03', '王五' );
CREATE TABLE SC ( SId VARCHAR ( 10 ), CID VARCHAR ( 10 ), score DECIMAL ( 18, 1 ) );
INSERT INTO SC
VALUES
 ( '01', '01', 80 );
INSERT INTO SC
VALUES
 ( '01', '02', 90 );
INSERT INTO SC
VALUES
 ( '01', '03', 99 );
INSERT INTO SC
VALUES
 ( '02', '01', 70 );
INSERT INTO SC
VALUES
 ( '02', '02', 60 );
INSERT INTO SC
VALUES
 ( '02', '03', 80 );
INSERT INTO SC
VALUES
 ( '03', '01', 80 );
INSERT INTO SC
VALUES
 ( '03', '02', 80 );
INSERT INTO SC
VALUES
 ( '03', '03', 80 );
INSERT INTO SC
VALUES
 ( '04', '01', 50 );
INSERT INTO SC
VALUES
 ( '04', '02', 30 );
INSERT INTO SC
VALUES
 ( '04', '03', 20 );
INSERT INTO SC
VALUES
 ( '05', '01', 76 );
INSERT INTO SC
VALUES
 ( '05', '02', 87 );
INSERT INTO SC
VALUES
 ( '06', '01', 31 );
INSERT INTO SC
VALUES
 ( '06', '03', 34 );
INSERT INTO SC
VALUES
 ( '07', '02', 89 );
INSERT INTO SC
VALUES
 ( '07', '03', 98 );
  • MySql:
create table Student(sid varchar(10),sname varchar(10),sage datetime,ssex nvarchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
create table Course(cid varchar(10),cname varchar(10),tid varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
create table Teacher(tid varchar(10),tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
create table SC(sid varchar(10),cid varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);