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  • SQL语法练习(一)

    1. 查询学习课程”python”比课程 “java” 成绩高的学生的学号;
      – 思路：
      – 获取所有有python课程的人（学号，成绩） - 临时表
      – 获取所有有java课程的人（学号，成绩） - 临时表
      – 根据学号连接两个临时表：
      – 学号 | 物理成绩 | 生物成绩
      – 然后再进行筛选
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      select A.s_id from
      		(select s_id, num as python from score left join course on score.c_id = course.c_id where course.c_name = 'python') as A
      		left join
      		(select s_id, num as java   from score left join course on score.c_id = course.c_id where course.c_name = 'java')   as B
      		on A.s_id = B.s_id where A.python > B.java;

    1. 查询平均成绩大于65分的同学的姓名和平均成绩(保留两位小数);

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      SELECT student.s_name as names, round(AVG(score.num), 2) as average
      FROM student, score
      WHERE student.s_id = score.s_id
      GROUP BY student.s_name
      HAVING AVG(score.num) > 65;

    2. 查询所有同学的姓名、选课数、总成绩

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      SELECT student.s_name, COUNT(score.s_id) as course_num, SUM(score.num) as total_grades
      FROM student, score
      WHERE student.s_id = score.s_id
      GROUP BY student.s_name
      ORDER BY student.s_name;

    3. 查询所有的课程的名称以及对应的任课老师姓名;

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      SELECT course.c_name, teacher.t_name  
      FROM course, teacher
      WHERE course.t_id = teacher.t_id;

    4. 查询没学过“alex”老师课的同学的姓名;

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      SELECT s_name
      FROM student
      WHERE student.s_id NOT IN (
      			SELECT DISTINCT score.s_id
      			FROM score, course, teacher
      			WHERE course.c_id = score.c_id AND teacher.t_id = course.t_id AND teacher.t_name = 'alex'
      			)

    5. 查询学过’python’并且也学过编号’java’课程的同学的姓名

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as cid
      		 FROM score, course
      		 WHERE score.c_id = course.c_id AND (course.c_name = 'python' OR course.c_name = 'java')) as B
      LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) > 1;

    6. 查询学过“alex”老师所教的全部课程的同学的姓名

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as cid
      		 FROM score, course, teacher
      		 WHERE score.c_id = course.c_id AND teacher.t_id = course.t_id AND teacher.t_name = 'alex') as B
          LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) = (SELECT COUNT(course.c_id) FROM
      												course, teacher
      												WHERE
      												teacher.t_id = course.t_id AND teacher.t_name='alex');

    7. 查询挂科超过两门(包括两门)的学生姓名;

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as大专栏  SQL语法练习(一)span> cid
      		 FROM score, course
      		 WHERE score.c_id = course.c_id  AND (score.num < 60 OR score.num ISNULL)) as B
           LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) > 1;

    8. 查询有课程成绩小于60分的同学的姓名;

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as cid
      		 FROM score, course
      		 WHERE score.c_id = course.c_id  AND score.num < 60) as B
           LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name;

    9. 查询选修了全部课程的学生姓名;

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as cid
      		 FROM score, course
      		 WHERE score.c_id = course.c_id) as B
           LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) = (SELECT COUNT(*) FROM course);


      SELECT s_name FROM
      student
      LEFT JOIN score ON student.s_id = score.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) = (SELECT COUNT(*) FROM course)

    10. 查询至少有一门课程与“貂蝉”同学所学课程相同的同学姓名；

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      SELECT s_name
      FROM student, score
      WHERE student.s_id = score.s_id AND student.s_name != '貂蝉' AND score.c_id IN (SELECT c_id FROM score, student
      																									 WHERE score.s_id = student.s_id AND student.s_name = '貂蝉')

      GROUP BY s_name;

    11. 查询学过’貂蝉’同学全部课程的其他同学姓名；

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      SELECT s_name FROM
      		(SELECT score.s_id as sid, score.c_id as cid
      		 FROM score, course, student
      		 WHERE
      		     score.c_id = course.c_id AND
      				 student.s_id = score.s_id AND
      				 student.s_name != '貂蝉' AND
      				 score.c_id in (SELECT c_id FROM score, student WHERE student.s_name = '貂蝉' AND student.s_id = score.s_id)
      		) as B
      LEFT JOIN student ON B.sid = student.s_id
      GROUP BY s_name
      HAVING COUNT(s_name) >= (SELECT COUNT(score.c_id) FROM score, student WHERE student.s_name = '貂蝉' AND student.s_id = score.s_id);

    – ——————————————————————————————————————————————————————————————
    – INSERT INTO student VALUES (1, ‘鲁班’, 12, ‘男’);
    – INSERT INTO student VALUES (2, ‘貂蝉’, 20, ‘女’);
    – INSERT INTO student VALUES (3, ‘刘备’, 35, ‘男’);
    – INSERT INTO student VALUES (4, ‘关羽’, 34, ‘男’);

    – INSERT INTO student VALUES (5, ‘张飞’, 33, ‘女’);

    –

    –
    – INSERT INTO teacher VALUES (1, ‘大王’);
    – INSERT INTO teacher VALUES (2, ‘alex’);
    – INSERT INTO teacher VALUES (3, ‘egon’);

    – INSERT INTO teacher VALUES (4, ‘peiqi’);

    – INSERT INTO course VALUES (1, ‘python’, 1);
    – INSERT INTO course VALUES (2, ‘java’, 2);
    – INSERT INTO course VALUES (3, ‘linux’, 3);
    – INSERT INTO course VALUES (4, ‘go’, 2);

    – INSERT INTO score VALUES (1, 1, 1, 79);
    – INSERT INTO score VALUES (2, 1, 2, 77);
    – INSERT INTO score VALUES (3, 1, 3, 58);
    – INSERT INTO score VALUES (4, 2, 2, 66);
    – INSERT INTO score VALUES (5, 2, 3, 77);
    – INSERT INTO score VALUES (6, 3, 1, 61);
    – INSERT INTO score VALUES (7, 3, 2, 64);

    – INSERT INTO score VALUES (8, 4, 3, 70);
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  • 原文地址：https://www.cnblogs.com/lijianming180/p/12286159.html
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