数据库【mysql篇】典型的一些练习题目

CREATE TABLE class (
  cid     INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
  caption VARCHAR(20)
)
  DEFAULT CHARSET = ‘utf8‘;

CREATE TABLE student (
  sid      INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
  sname    VARCHAR(20),
  gender   VARCHAR(20),
  class_id INT,
  CONSTRAINT fk_clsid FOREIGN KEY (class_id) REFERENCES class (cid)
)
  DEFAULT CHARSET = ‘utf8‘;

CREATE TABLE teacher (
  tid   INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
  tname VARCHAR(32)
)
  DEFAULT CHARSET = ‘utf8‘;

CREATE TABLE course (
  cid       INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
  cname     VARCHAR(20),
  tearch_id INT,
  CONSTRAINT fk_tea FOREIGN KEY (tearch_id) REFERENCES teacher (tid)
)
  DEFAULT CHARSET = ‘utf8‘;

CREATE TABLE score (
  sid        INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
  student_id INT,
  corse_id   INT,
  number     INT,
  CONSTRAINT fk_sco_stu FOREIGN KEY (student_id) REFERENCES student (sid),
  CONSTRAINT fk_sco_cor FOREIGN KEY (corse_id) REFERENCES course (cid)
)
  DEFAULT CHARSET = ‘utf8‘;

# 查询“生物”课程比“物理”课程成绩高的所有学生的学号；
SELECT tb1.student_id
FROM (SELECT
        student_id,
        number
      FROM score s LEFT JOIN course c ON s.corse_id = c.cid
      WHERE c.cname = ‘体育‘) AS tb1 LEFT JOIN (
                                               SELECT
                                                 student_id,
                                                 number
                                               FROM score s LEFT JOIN course c ON s.corse_id = c.cid
                                               WHERE c.cname = ‘物理‘) AS tb2 ON tb1.student_id = tb2.student_id
WHERE tb1.number > tb2.number;

# 查询平均成绩大于60分的同学的学号和平均成绩；
SELECT
  s.sid,
  avg(number)
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
GROUP BY s.sid
HAVING avg(number) > 60;

# 查询所有同学的学号、姓名、选课数、总成绩；
SELECT
  student.sid,
  student.sname,
  count(s.corse_id),
  sum(s.number)
FROM student
  LEFT JOIN score s ON student.sid = s.student_id
GROUP BY student.sid;

# 查询姓“李”的老师的个数；
SELECT count(tid)
FROM teacher
WHERE tname LIKE "李%";

# 查询没学过“叶平”老师课的同学的学号、姓名；
SELECT
  sid,
  sname
FROM student
WHERE sid NOT IN (
  SELECT DISTINCT student_id
  FROM score
  WHERE corse_id IN (
    SELECT course.cid
    FROM course
      LEFT JOIN teacher t ON course.tearch_id = t.tid
    WHERE t.tname = "叶平"
  ));

# 查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
SELECT
  tb2.sid,
  tb2.sname
FROM
  (SELECT
     student_id,
     corse_id
   FROM score
   WHERE corse_id = 2 OR corse_id = 3) AS tb1
  LEFT JOIN student tb2 ON tb2.sid = tb1.student_id
GROUP BY student_id
HAVING count(student_id) > 1;

# 查询学过“叶平”老师所教的所有课的同学的学号、姓名；
SELECT
  student.sid,
  student.sname
FROM student
WHERE sid IN (
  SELECT DISTINCT student_id
  FROM score
  WHERE corse_id IN (
    SELECT course.cid
    FROM course
      LEFT JOIN teacher t ON course.tearch_id = t.tid
    WHERE t.tname = ‘苍空‘)
);

# 9.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；
# 跟第一题差不多

# 查询有课程成绩小于60分的同学的学号、姓名；
SELECT
  student.sid,
  student.sname
FROM student
WHERE sid IN (
  SELECT DISTINCT student_id
  FROM score
  WHERE number < 60
);

# 查询没有学全所有课的同学的学号、姓名；
SELECT
  sid,
  sname
FROM student
WHERE sid IN (
  SELECT student_id
  FROM score
  GROUP BY student_id
  HAVING count(number) = (
    SELECT COUNT(1)
    FROM course));

SELECT
  s.sid,
  s.sname
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
GROUP BY score.student_id
HAVING count(number) = (SELECT count(1)
                        FROM course);

# 查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；
#    1 先找到001同学所学的所有课程id
#    2.条件学号不为1 以及所学课程in 1号学生所学的id里面
SELECT
  s.sid,
  s.sname
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
WHERE student_id != 1
      AND score.corse_id IN (
  SELECT corse_id
  FROM score
  WHERE student_id = 1)
GROUP BY student_id;

# 查询至少学过学号为“002”同学所有课的其他同学学号和姓名；
SELECT
  student_id,
  sname,
  count(score.corse_id)
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
WHERE score.student_id != 2 AND score.corse_id IN (
  SELECT corse_id
  FROM score
  WHERE student_id = 2)
GROUP BY student_id
HAVING count(corse_id) = (SELECT count(corse_id)
                          FROM score
                          WHERE student_id = 2);


SELECT
  student_id,
  sname,
  count(corse_id)
FROM score
  LEFT JOIN student ON score.student_id = student.sid
WHERE student_id != 1 AND corse_id IN
                          (SELECT corse_id
                           FROM score
                           WHERE student_id = 1)
GROUP BY student_id
HAVING count(corse_id) =
       (SELECT count(corse_id)
        FROM score
        WHERE student_id = 1);

# 查询和“001”号的同学学习的课程完全相同的其他同学学号和姓名；
# 总个数相等  但 不一定所学的就等于1号所学的
# 2号所学的课程都被学到  但验证不了总个数相等
SELECT
  student_id,
  sname
FROM score
  LEFT JOIN student ON score.student_id = student.sid
# 总的数量=1号总的数量
WHERE student_id IN (SELECT student_id
                     FROM score
                     WHERE student_id != 2
                     GROUP BY student_id
                     HAVING count(corse_id) = (SELECT count(1)
                                               FROM score
                                               WHERE student_id = 2))
      AND corse_id IN (
  # 1号所学的课程数量都已被学到
  SELECT corse_id
  FROM score
  WHERE corse_id IN (SELECT corse_id
                     FROM score
                     WHERE student_id = 2)
  GROUP BY student_id
  HAVING count(corse_id) = (SELECT count(*)
                            FROM score
                            WHERE student_id = 2)
);

# 15、删除学习“叶平”老师课的score表记录；
DELETE FROM score
WHERE score.corse_id IN
      (SELECT cid
       FROM course
         LEFT JOIN teacher t ON course.tearch_id = t.tid
       WHERE t.tname = ‘叶平‘);

# 向SC表中插入一些记录，这些记录要求符合以下条件：
# ①没有上过编号“002”课程的同学学号；
# ②插入“002”号课程的平均成绩；
INSERT INTO score (student_id, corse_id, number)
  SELECT
    sid,
    2,
    (SELECT avg(number)
     FROM score
     WHERE corse_id = 2)
  FROM
    student
  WHERE sid NOT IN (
    SELECT student_id
    FROM score
    WHERE corse_id = 2
  );

# 17、按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；
SELECT
  student_id,
  (SELECT number
   FROM score
     LEFT JOIN course c ON score.corse_id = c.cid
   WHERE c.cname = ‘生物‘ AND score.student_id = sc.student_id) AS "生物",
  (SELECT number
   FROM score
     LEFT JOIN course c ON score.corse_id = c.cid
   WHERE c.cname = ‘物理‘ AND score.student_id = sc.student_id) AS "物理",
  (SELECT number
   FROM score
     LEFT JOIN course c ON score.corse_id = c.cid
   WHERE c.cname = ‘体育‘ AND score.student_id = sc.student_id) AS "体育",
  count(sc.corse_id),
  avg(number)
FROM score AS sc
WHERE sc.corse_id IN (SELECT cid
                      FROM course
                      WHERE course.cname = ‘生物‘ OR course.cname = ‘物理‘ OR course.cname = ‘体育‘)
GROUP BY sc.student_id
ORDER BY avg(number);

# 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
  corse_id,
  c.cname,
  max(number),
  min(number)
FROM score
  LEFT JOIN course c ON score.corse_id = c.cid
GROUP BY corse_id;

# 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；
# 思路：case when .. then
SELECT
  corse_id,
  avg(number)                      AS avgnum,
  sum(CASE WHEN score.number > 60
    THEN 1
      ELSE 0 END) / count(1) * 100 AS percent
FROM score
GROUP BY corse_id
ORDER BY avgnum ASC, percent DESC;

# 20、课程平均分从高到低显示（显示任课老师）
SELECT
  avg(number),
  t.tname
FROM score
  LEFT JOIN course c ON score.corse_id = c.cid
  LEFT JOIN teacher t ON c.tearch_id = t.tid
GROUP BY corse_id
ORDER BY avg(number) DESC;

# 21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
# 思路 先找到第一名和第四名的值
SELECT
  score.sid,
  score.corse_id,
  score.number,
  T.first_num,
  T.second_num
FROM score
  LEFT JOIN
  (SELECT
     sid,
     (SELECT number
      FROM score AS s2
      WHERE s2.corse_id = s1.corse_id
      ORDER BY number DESC
      LIMIT 0, 1) AS first_num,
     (SELECT number
      FROM score AS s2
      WHERE s2.corse_id = s1.corse_id
      ORDER BY number DESC
      LIMIT 3, 1) AS second_num,
   FROM score
     AS s1)
    AS T ON score.sid = T.sid
WHERE score.number <= T.first_num AND score.number >= T.second_num;

# 22、查询每门课程被选修的学生数；
SELECT
  corse_id,
  count(1)
FROM score
GROUP BY corse_id;

# 23、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,
  count(1)
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
GROUP BY student_id
HAVING count(1) = 1;

# 24、查询男生、女生的人数；
# 男生总数为一张表 女生总数为一张表，每张表里面都只有一个字段
# 查询两张表
SELECT *
FROM
  (SELECT count(1) AS man
   FROM student
   WHERE student.gender = ‘男‘) AS A,
  (SELECT count(1) AS wuman
   FROM student
   WHERE student.gender = ‘女‘) AS B

# 25、查询姓“张”的学生名单；
SELECT *
FROM student
WHERE sname LIKE ‘张%‘;

# 26、查询同名同姓学生名单，并统计同名人数；
SELECT
  sname,
  count(1)
FROM student
GROUP BY sname;

# 27、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
  corse_id,
  avg(if(isnull(number), 0, number)) AS avg
FROM score
GROUP BY corse_id
ORDER BY avg ASC, corse_id DESC

# 28、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
  s.sid,
  s.sname,
  avg(if(isnull(number), 0, number)) AS avg
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
GROUP BY student_id
HAVING avg > 85;

# 29、查询课程名称为“数学”，且分数低于60的学生姓名和分数；
SELECT
  s.sid,
  s.sname,
  score.number
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
  LEFT JOIN course c ON score.corse_id = c.cid
WHERE c.cname = ‘数学‘ AND number < 60;

# 30、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
  s.sid,
  s.sname,
  score.number,
  score.corse_id
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
WHERE corse_id = ‘3‘ AND number > 80;

# 31、求选了课程的学生人数
# 第一种做法
SELECT count(DISTINCT student_id)
FROM score;

# 第二种做法
SELECT count(c)
FROM (
       SELECT count(student_id) AS c
       FROM score
       GROUP BY student_id) AS A;

# 查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩；
# 思路 先找到杨艳所教的课程id，然后再根据课程id分组 排序取第一个
SELECT
  corse_id,
  s.sname,
  score.number
FROM score
  LEFT JOIN student s ON score.student_id = s.sid
WHERE score.corse_id IN (SELECT course.cid
                         FROM course
                           LEFT JOIN teacher t ON course.tearch_id = t.tid
                         WHERE t.tname = ‘波多‘)
ORDER BY number DESC
LIMIT 1;

# 33、查询各个课程及相应的选修人数；
SELECT
  corse_id,
  count(1),
  c.cname
FROM score
  LEFT JOIN course c ON score.corse_id = c.cid
GROUP BY corse_id;

# 34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩；
# 同一张表进行链表操作
SELECT DISTINCT
  s1.corse_id,
  s2.corse_id,
  s1.number,
  s2.number,
  s1.student_id,
  s2.student_id
FROM score AS s1, score AS s2
WHERE s1.corse_id != s2.corse_id AND s1.number = s2.number

# 35、查询每门课程成绩最好的前两名；
# 思路 先找到第一名跟第二名同学的成绩 组成一张新的表

SELECT
  score.sid,
  corse_id,
  score.number
FROM score
  LEFT JOIN
  (SELECT
     sid,
     (SELECT number
      FROM score AS s2
      WHERE s2.corse_id = s1.corse_id
      ORDER BY number DESC
      LIMIT 1 OFFSET 0) AS first_num,
     (SELECT number
      FROM score AS s2
      WHERE s2.corse_id = s1.corse_id
      ORDER BY number DESC
      LIMIT 1 OFFSET 1) AS second_num
   FROM score AS s1)
    AS T
    ON score.sid = T.sid
WHERE score.number <= T.first_num AND score.number >= T.second_num
ORDER BY score.corse_id DESC, score.number DESC;

# 36、检索至少选修两门课程的学生学号；
SELECT score.sid
FROM score
GROUP BY student_id
HAVING count(student_id) > 1;

# 37、查询全部学生都选修的课程的课程号和课程名；
SELECT
  cid,
  course.cname
FROM course
WHERE course.cid IN
      (SELECT corse_id
       FROM score
       GROUP BY corse_id
       HAVING count(1) = (SELECT count(1)
                          FROM student));

# 38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；
# 先找到叶平老师所教的课程id
# 然后找到有学过任意一门是属于叶平老师的课的学生id
# 然后学生不在里面
SELECT student.sname
FROM student
WHERE sid NOT IN (
  SELECT student_id
  FROM score
  WHERE score.corse_id IN (
    SELECT cid
    FROM course
      LEFT JOIN teacher ON course.tearch_id = teacher.tid
    WHERE tname = ‘苍空‘
  )
);

# 错误的做法
# select student_id,student.sname from score
#   left join student on score.student_id = student.sid
#   where score.corse_id not in (
#       select cid from course left join teacher on course.tearch_id = teacher.tid where tname = ‘张磊老师‘
#   )
#   group by student_id

# 39、查询两门以上不及格课程的同学的学号及其平均成绩；
SELECT student_id,count(1) FROM score WHERE number < 60
GROUP BY student_id
HAVING count(1) > 2;

# 40、检索“004”课程分数小于60，按分数降序排列的同学学号；
SELECT student_id,number FROM score WHERE number < 60 and corse_id = 4
ORDER BY number DESC;

# 41、删除“002”同学的“001”课程的成绩；
# SELECT * FROM score WHERE student_id = 2 and corse_id = 1;
DELETE FROM score WHERE student_id = 2 and corse_id = 1;