MySql经典语句练习50题 ---- 1 ~ 9题
目录
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数：
2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数：
3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩：
4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 (包括有成绩的和无成绩的)：
5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩：
6、查询"李"姓老师的数量：
7、查询学过"张三"老师授课的同学的信息：
8、查询没学过"张三"老师授课的同学的信息：
9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息：
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数：
SELECT student.*, sc1.s_score score1, sc2.s_score score2
FROM student, score sc1, score sc2
WHERE student.s_id = sc1.s_id AND sc1.s_id = sc2.s_id AND sc1.c_id = '01' AND sc2.c_id = '02' AND sc1.s_score > sc2.s_score;2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数：
SELECT student.*, sc1.s_score score1, sc2.s_score score2
FROM student, score sc1, score sc2
WHERE student.s_id = sc1.s_id AND sc1.s_id = sc2.s_id AND sc1.c_id = '01' AND sc2.c_id = '02' AND sc1.s_score < sc2.s_score;3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩：
SELECT student.s_id, s_name, avg(s_score) avg
FROM student, score
WHERE student.s_id = score.s_id
GROUP BY student.s_id
HAVING AVG(s_score) >= 60;4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 (包括有成绩的和无成绩的)：
SELECT student.s_id, s_name, AVG(s_score) avg
FROM student LEFT JOIN score ON (student.s_id = score.s_id)
GROUP BY student.s_id
HAVING AVG(s_score) < 60 OR AVG(s_score) IS NULL;5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩：
SELECT student.s_id, s_name, count(c_id) number, sum(s_score) sum
FROM student, score
WHERE student.s_id = score.s_id
GROUP BY student.s_id;6、查询"李"姓老师的数量：
SELECT count(t_name) number
FROM teacher
WHERE t_name LIKE '李%';7、查询学过"张三"老师授课的同学的信息：
SELECT *
FROM student
WHERE EXISTS (SELECT *
FROM score, teacher
WHERE student.s_id = score.s_id AND c_id = t_id and t_name = '张三');8、查询没学过"张三"老师授课的同学的信息：
SELECT *
FROM student
WHERE NOT EXISTS (SELECT *
FROM score, teacher
WHERE student.s_id = score.s_id AND c_id = t_id and t_name = '张三');9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息：
SELECT student.*
FROM student, score sc1, score sc2
WHERE student.s_id = sc1.s_id and student.s_id = sc2.s_id and sc1.c_id = '01' and sc2.c_id = '02';