sql语句练习50题(含解析)

1、查询”01″课程比”02″课程成绩高的学生的信息及课程分数(偏难)

SELECT
	st.*,
	sc1.s_score 01_score,
	sc2.s_score 02_score
FROM
	student st,
	score sc1,
	score sc2
WHERE
	st.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

通过score表自连接后，过滤出每位学生 “01”课程分数和 “02”课程分数的一条记录，最后将01”课程分数大于 “02”课程分数记录筛选出来

2、查询”01″课程比”02″课程成绩低的学生的信息及课程分数

与第1题思路一样

3、查询平均成绩大于等于60分的每个同学的学生编号和学生姓名和平均成绩

SELECT
	st.s_id,
	s_name,
	avg(sc.s_score)
FROM
	student st,
	score sc
WHERE
	st.s_id = sc.s_id
GROUP BY
	st.s_id,
	st.s_name
HAVING
	avg(sc.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
	st.s_id,
	s_name,
	ROUND(avg(sc.s_score),2) avgsc
FROM
	student st LEFT JOIN
	score sc
on
	st.s_id = sc.s_id
GROUP BY
	st.s_id,
	st.s_name
HAVING
	avg(sc.s_score)  60 or avgsc 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
	st.s_id,
	st.s_name,
	count(sc.c_id) AS sum_course,
	sum(sc.s_score) AS sum_score
FROM
	student st
LEFT JOIN score sc ON st.s_id = sc.s_id
GROUP BY
	st.s_id,
	st.s_name

此题比较简单，使用分组即可。需要注意下在mysql中使用group by后,select后出现的字段要么是
分组的字段要么是聚集函数不然会报错

6、查询”李”姓老师的数量

SELECT
	count(*)
FROM
	teacher
WHERE
	t_name LIKE '李%'

这一题就是考察 mysql中like关键字的使用

7、查询学过”张三”老师授课的同学的信息

SELECT
	st.*
FROM
	student st
WHERE
	st.s_id IN (
		SELECT
			s_id
		FROM
			score
		WHERE
			c_id = (
				SELECT
					c_id
				FROM
					course
				WHERE
					t_id = (
						SELECT
							t_id
						FROM
							teacher
						WHERE
							t_name = '张三'
					)
			)
	)

这题我使用的是where型子查询需要使用到in关键字，使用连接查询也可以.根据表之间的连接关系:

1. 首先在teacher表中通过t_name拿到t_id
2. 在course表中通过t_id拿到c_id
3. 在course表中通过c_id拿到s_id
4. 最后在student表中通过s_id拿到学过‘张三’老师授课的学生信息

类似这种子查询都可以用连接查询替代

8、查询没学过”张三”老师授课的同学的信息

此题只需要把上一题中的in关键改成not in关键字即可

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
	*
FROM
	student
WHERE
	s_id IN (
		SELECT
     sc1.s_id
		FROM
			score sc1
		JOIN score sc2 ON sc1.s_id = sc2.s_id
		WHERE
			sc1.c_id = '01'
		AND sc2.c_id = '02'
	)

这一题的方法与第一题类似，第一题搞懂了，这一题就很简单了。直接通过score表进行自连结运算然后筛选出即选修了’01’号课程又选修了’02’号课程的s_id

10、查询学过编号为”01″但是没有学过编号为”02″的课程的同学的信息(偏难)

SELECT
	st.* 
FROM
	student st 
WHERE
	st.s_id IN ( SELECT s_id FROM score WHERE c_id = '01' ) 
	AND st.s_id NOT IN ( SELECT s_id FROM score WHERE c_id = '02' )

11、查询没有学全所有课程的同学的信息

SELECT
	*
FROM
	student
WHERE
	s_id NOT IN (
		SELECT
			s_id
		FROM
			score
		GROUP BY
			s_id
		HAVING
			count(c_id) = (
				SELECT
					count(c_id)
				FROM
					course
			)
	)

12、查询至少有一门课与学号为”01″的同学所学相同的同学的信息

SELECT DISTINCT
	st.*
FROM
	student st
LEFT JOIN score sc ON st.s_id = sc.s_id
GROUP BY
	s_id,
	c_id
HAVING
	c_id IN (
		SELECT
			c_id
		FROM
			score
		WHERE
			s_id = '01'
	)

SELECT DISTINCT st.* FROM student st LEFT JOIN score sc ON st.s_id=sc.s_id WHERE c_id IN (
SELECT c_id FROM score WHERE s_id='01')

select * from Student where s_id in (SELECT DISTINCT s_id from Score where c_id in (select c_id from Score where s_id='01'))

这题关键在于用其他每个学生所学课程号与”01″学生所学课程的课程号作对比

13、查询和”01″号的同学学习的课程完全相同的其他同学的信息

SELECT
	st.*
FROM
	student st
JOIN score sc ON st.s_id = sc.s_id
WHERE
	sc.c_id IN (
		SELECT
			c_id
		FROM
			score
		WHERE
			s_id = '01'
	)
AND st.s_id != '01'
GROUP BY
	st.s_id
HAVING
	count(st.s_id) = (
		SELECT
			count(*)
		FROM
			score
		WHERE
			s_id = '01'
	)

select  Student.* from Student left JOIN Score on Student.s_id = Score.s_id where Score.c_id in (select c_id from Score where s_id='01') and Score.s_id  '01' GROUP BY Student.s_id HAVING COUNT(*)=(select count(*) from Score where s_id='01')

14、查询没学过”张三”老师讲授的任一门课程的学生姓名

SELECT
	*
FROM
	student
WHERE
	s_id NOT IN (
		SELECT
			st.s_id
		FROM
			student st,
			score sc
		WHERE
			st.s_id = sc.s_id
		AND c_id IN (
			SELECT
				c_id
			FROM
				teacher te
			LEFT JOIN course c ON te.t_id = c.t_id
			WHERE
				t_name = '张三'
		)
	)

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

SELECT
	st.s_id,
	st.s_name,
	round(avg(sc.s_score))
FROM
	student st,
	score sc
WHERE
	st.s_id = sc.s_id
AND s_score  60
GROUP BY
	s_id
HAVING
	count(*) >= 2

16、检索”01″课程分数小于60，按分数降序排列的学生信息

SELECT
	st.*,
	sc.c_id,
	sc.s_score
FROM
	student st,
	score sc
WHERE
	st.s_id = sc.s_id
AND s_score  60
AND c_id = '01'
ORDER BY
	s_score DESC

17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select st.s_id,
(select s_score from score where c_id='01' and s_id=st.s_id) as '语文' ,
(select s_score from score where c_id='02' and s_id=st.s_id) as '数学',
(select s_score from score where c_id='03' and s_id=st.s_id) as  '英语',
round((select avg(s_score) from score where s_id=st.s_id group by s_id),2) as '平均分'
from student st order by  平均分 desc

这一题用到了子查询的另一种方式即创建计算字段，该子查询对检索出的每个student执行一次，在此例中，该查询共执行了8次，因为检索出8名学生，另外当在group by 或 order by后面使用中文时，不要加单引号，否则不生效

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，中等率，优良率，优秀率及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

explain select sc.c_id,c_name,max(s_score),min(s_score),ROUND(avg(s_score),2) '平均分',
ROUND(100*(SUM(case when sc.s_score>=60 then 1 else 0 end)/SUM(case when sc.s_score then 1 else 0 end)),2) as '及格率',
ROUND(100*(SUM(case when sc.s_score>=70 and sc.s_score80 then 1 else 0 end)/SUM(case when sc.s_score then 1 else 0 end)),2) as '中等率',
ROUND(100*(SUM(case when sc.s_score>=80 and sc.s_score90 then 1 else 0 end)/SUM(case when sc.s_score then 1 else 0 end)),2) as '优良率',
ROUND(100*(SUM(case when sc.s_score>=90 then 1 else 0 end)/SUM(case when sc.s_score then 1 else 0 end)),2) as '优秀率'
from course c,score sc where c.c_id=sc.c_id group by sc.c_id

19、按各科成绩进行排序，并显示排名

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

SELECT c.t_id,t.t_name,s.c_id,ROUND(AVG(s_score),2) avgsc FROM teacher t,course c,score s WHERE t.t_id=c.t_id AND c.c_id=s.c_id GROUP BY s.c_id,c.t_id,t.t_name ORDER BY avgsc DESC

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

select c_id ,count(*) from score group by c_id 

27、查询出只有两门课程的全部学生的学号和姓名

select st.s_id,st.s_name,count(*) '课程数' from student st left join score sc on st.s_id=sc.s_id group by s_id having count(*)=2

select s_id,s_name from Student WHERE s_id in (select s_id  FROM Score GROUP BY s_id HAVING count(*) = 2)

28、查询男生、女生人数

select s_sex,(select count(s_sex) from student st2 where s_sex = st1.s_sex) '人数' from student st1 group by s_sex

select  DISTINCT s_sex,(select count(s_sex) from Student st2 where s_sex = st1.s_sex) '人数' from Student st1 

29、查询名字中含有”风”字的学生信息

 select * from student where s_name like '%风%'

30、查询同名同性学生名单，并统计同名人数

select s_name,count(*)-1 '同名人数' from student group by s_name having count(*)>=2

31、查询1990年出生的学生名单

select s_name from student where s_birth like '1990%'

32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select c_id, ROUND(avg(s_score),2) '平均成绩' from score group by c_id order by 平均成绩 desc ,c_id

33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

select st.s_id,s_name,ROUND(avg(s_score),2) '平均成绩' from student st,score sc where st.s_id=sc.s_id group by st.s_id , s_name having avg(s_score)>=85

select s.s_id,s.s_name,ROUND(AVG(sc.s_score)) '平均分' from Student s LEFT JOIN Score sc ON s.s_id=sc.s_id GROUP BY s.s_id,s.s_name HAVING AVG(sc.s_score)>=85

34、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_name,s_score from student st,course c,score sc where st.s_id=sc.s_id and sc.c_id=c.c_id and c_name='数学' and s_score60

select  distinct s_name,s_score from Student s LEFT JOIN Score sc on s.s_id=sc.s_id LEFT JOIN Course c on c.c_id=sc.c_id where c_name='数学' and s_score60

35、查询所有学生的课程及分数情况

select st.s_id,s_name,c_name,s_score from student st,course c,score sc where st.s_id=sc.s_id and sc.c_id=c.c_id

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 s_name,c_name,s_score from student st,course c,score sc where st.s_id = sc.s_id and c.c_id =sc.c_id and s_score>70

37、查询不及格的课程

select st.*,c_name,s_score from student st,score sc,course c where st.s_id=sc.s_id and sc.c_id=c.c_id and s_score60

38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名

select st.s_id,st.s_name from student st,score sc where st.s_id=sc.s_id and c_id='01' and s_score>=80

39、求每门课程的学生人数

select c.c_name '课程名', count(*) '人数' from Course c,Score s where c.c_id=s.c_id GROUP BY s.c_id,c.c_name

select c_name '课程名' ,(select count(*) from Score s1 where c_id=c.c_id) '人数' from Course c

40、查询选修”张三”老师所授课程的学生中，成绩最高的学生信息及其成绩

select st.*,s_score from student st,score sc where st.s_id=sc.s_id and c_id=(select c_id from teacher te left join  course c on te.t_id=c.t_id where t_name='张三') order by s_score desc limit 1

41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩

select distinct s1.* FROM score s1 ,score s2 where s1.c_id !=s2.c_id and s1.s_score=s2.s_score

42、查询每门功成绩最好的前两名(偏难)

select sc1.s_id,sc1.c_id,sc1.s_score from score sc1
		where (select COUNT(1) from score sc2 where sc2.c_id=sc1.c_id and sc2.s_score>=sc1.s_score)2 ORDER BY sc1.c_id

首先，select count(1)表示查询出表中符合条件的行数；
sc2.c_id=sc1.c_id and sc2.s_score>=sc1.s_score表示查询条件；
select COUNT(1) from score sc2 where sc2.c_id=sc1.c_id and sc2.s_score>=sc1.s_score总体的意思就是从表sc2中查询出满足sc2.c_id=sc1.c_id and sc2.s_score>=sc1.s_score条件的行数；
结合完整的sql语句来看，这个查询出的行数要 这样这个语句简单点理解就是：从sc1表查询sc1.s_id,sc1.c_id,sc1.s_score这三列，查询条件是”行数

43、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 c_id,count(*) '选修人数' from score group by c_id having count(*)>5 order by 选修人数 desc ,c_id

44、检索至少选修两门课程的学生学号

select s_id from score group by s_id having count(*)>=2

45、查询选修了全部课程的学生信息

select st.* from student st,score sc where st.s_id=sc.s_id  group by sc.s_id having count(*) = (select count(*) from course)