SQL查询语句练习（40道）

素材题目来源

50道SQL练习题及答案与详细分析偶尔在网上看到了博主在挑战自己的博客，我想着自己也挑战一下自己,毕竟上学期才学完数据库，现在应该也是忘得差不多了，这里记录一下，如有记忆有点混乱的就做一下笔记，没有问题的就不去添加说明了（大胆的想自己已经是掌握好了的）。部分的比较偏一点的（之前数据库课本上没有讲到的就用了他的答案），这个练习的是链接博主的前40道，因为后面的50道题目是使用函数的就没有必要的，关键的是锻炼自己的思维能力。

题目一

查询" 01 “课程比” 02 "课程成绩高的学生的信息及课程分数

使用exist语句直接查询，其实它可以和in语句相互替换

SELECT s1.Sid,Sname,Sage,Ssex,score
FROM student s1,sc sc1
WHERE s1.SId = sc1.SId AND sc1.CId='01' AND
EXISTS
(
SELECT *
FROM sc sc2
WHERE sc2.SId=s1.SId AND sc2.CId='02' AND sc2.score < sc1.score 
)

1.1 查询同时存在" 01 “课程和” 02 "课程的成绩情况

SELECT t1.SId,t1.CId AS '课程1',t1.score,t2.CId AS '课程2',t2.score  FROM
    (SELECT * FROM sc WHERE sc.CId = '01') AS t1, 
    (SELECT * FROM sc WHERE sc.CId = '02') AS t2
WHERE t1.SId = t2.SId

1.2查询存在" 01 “课程但可能不存在” 02 "课程的成绩情况(不存在时显示为 null )

SELECT * FROM
(SELECT * FROM sc where CId='01') as s1
LEFT JOIN
(SELECT * FROM sc WHERE CId='02') as s2
ON s1.SId = s2.SId

1.3查询不存在" 01 “课程但存在” 02 "课程的成绩情况

SELECT * FROM sc WHERE CId='02'
 AND SId NOT IN
(
SELECT SId FROM sc WHERE CId='01'
)

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.SId,sname,AVG(score)
FROM student,sc
WHERE student.SId=sc.SId
GROUP BY student.SId
HAVING AVG(score)>='60'

查询在 SC 表存在成绩的学生信息

SELECT DISTINCT student.* FROM student,sc WHERE student.SId=sc.SId

查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

SELECT student.SId, Sname,COUNT(*) AS '课程总数',SUM(score) AS '总成绩' 
FROM student,sc
WHERE student.SId=sc.SId
GROUP BY student.SId

4.1 查有成绩的学生信息

SELECT * FROM student
WHERE Sid IN 
(
SELECT SId FROM sc
)

查询「李」姓老师的数量

SELECT COUNT(*) FROM teacher WHERE Tname LIKE '李%'

查询学过「张三」老师授课的同学的信息

SELECT student.* FROM student,sc WHERE
student.SId=sc.SId AND CId IN
( 
SELECT sc.CId FROM sc,course,teacher WHERE
 sc.CId=course.CId AND course.TId=teacher.TId AND Tname='张三'
)

查询没有学全所有课程的同学的信息

这里使用了NOT IN的语句，然后判断“全部”是按照课程的数量判断的。与这样的“带有全称词”的查询，使用标准的是用双重NOT EXISTS语句（第9题）

SELECT * FROM student WHERE SId NOT IN
(
SELECT student.SId FROM student,sc WHERE student.Sid=sc.SId
GROUP BY student.SId HAVING COUNT(*)=
(
SELECT COUNT(*) FROM course
)
)

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 * FROM student s1 WHERE EXISTS
(
SELECT * FROM sc sc1 WHERE s1.SId=sc1.SId AND s1.SId<>'01' AND CId IN
(
SELECT CId FROM sc sc2 WHERE SId='01'
) 
)

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

当遇到这种全称量词修饰的查询语句的时候，我们可以把它转换成为相应的存在语句，即：

不存在这样的同学——他学的课程与‘01’号学生不存在相等的情况

SELECT * FROM student s1 WHERE s1.Sid<>'01' AND NOT EXISTS
(
SELECT * FROM sc sc1 WHERE sc1.SId='01' AND NOT EXISTS
(
SELECT * FROM sc sc2 WHERE sc2.SId=s1.SId AND sc1.CId=sc2.CId
)
)

查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT Sname FROM student WHERE NOT EXISTS
(
SELECT * FROM sc WHERE SId=student.SId AND CId IN
(
SELECT course.CId FROM course,teacher WHERE course.TId=teacher.TId AND Tname='张三'
)
)

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 student.SId, Sname,AVG(score) FROM student,sc WHERE student.SId=sc.SId AND student.SId IN
(
SELECT SId FROM sc WHERE score<'60' 
GROUP BY SId HAVING COUNT(*)>=2
)
GROUP BY student.SId, Sname

检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT * FROM student WHERE SId IN
(
SELECT SId FROM sc WHERE CId='01' AND score<'60' ORDER BY score DESC
)

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 * FROM sc LEFT JOIN (SELECT SId,AVG(score) AS avgscore FROM sc GROUP BY SId) as sc1
ON sc.SId=sc1.SId
ORDER BY avgscore DESC

注意:这里使用平均值进行排序的时候一定要为其声明一个别名，不然不能成功。

查询各科成绩最高分、最低分和平均分

SELECT CId, MAX(score) AS '最高分',MIN(score) AS '最低分',AVG(score) AS '平均分',COUNT(*)
FROM sc GROUP BY CId

按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺（待）

select a.cid, a.sid, a.score, count(b.score)+1 as rank
from sc as a 
left join sc as b 
on a.score<b.score and a.cid = b.cid
group by a.cid, a.sid,a.score
order by a.cid, rank ASC;

查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

SET @rank=0;
SELECT SId,sum,@rank := @rank+1 AS rank FROM
(
SELECT SId,SUM(score) AS sum FROM sc 
GROUP BY SId
ORDER BY SUM(score) DESC
)AS sc1

mysql中的变量的用法

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]，[60-0] 及所占百分比

select course.cname, course.cid,
sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) as "[100-85]",
sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) as "[85-70]",
sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) as "[70-60]",
sum(case when sc.score<=60 and sc.score>0 then 1 else 0 end) as "[60-0]"
from sc left join course
on sc.cid = course.cid
group by sc.cid;

知识点：左连接，case-when与集合函数的使用

查询各科成绩前三名的记录

select a.sid,a.cid,a.score from sc a 
left join sc b on a.cid = b.cid and a.score<b.score
group by a.cid, a.sid
having count(b.cid)<3
order by a.cid;

查询每门课程被选修的学生数

SELECT CId,COUNT(*) FROM sc GROUP BY CId

查询出只选修两门课程的学生学号和姓名

SELECT SId,Sname FROM student s WHERE EXISTS
(
SELECT * FROM sc WHERE s.SId=sc.SId GROUP BY SId HAVING COUNT(*)=2
)

查询男生、女生人数

SELECT Ssex,COUNT(*)
FROM student GROUP BY Ssex

查询名字中含有「风」字的学生信息

SELECT * FROM student WHERE Sname LIKE '%风%'

查询同名学生名单，并统计同名人数

SELECT s1.Sname,COUNT(*) FROM student s1,student s2 
WHERE s1.SId!=s2.SId AND s1.Sname=s2.Sname

查询 1990 年出生的学生名单

SELECT *
FROM student
WHERE YEAR(student.Sage)=1990;

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 CId,AVG(score) AS avg from sc 
GROUP BY CId
ORDER BY avg DESC

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 s.SId,s.Sname,avg(score) AS avgscore from student s,sc
WHERE s.SId=sc.SId GROUP BY s.SId HAVING avgscore>85

查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT Sname,score FROM student s,sc,course c
WHERE s.SId=sc.SId AND c.CId=sc.CId AND c.Cname='数学' AND score<'60'

查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

SELECT s.SId,s.Sname,score FROM student s LEFT JOIN sc
ON s.SId=sc.SId

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.SId, s.Sname,score,Cname FROM student s,sc,course c WHERE s.SId=sc.SId AND NOT EXISTS
(
SELECT * FROM sc WHERE sc.SId = s.SId AND score<'70'
)AND sc.CId=c.CId

查询存在不及格的课程

SELECT CId,Cname FROM course WHERE CId IN
(
SELECT CId FROM sc WHERE score<'60'
)

查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

SELECT SId, Sname FROM student WHERE SId IN
(
SELECT SId FROM sc WHERE CId='01' AND score>='80'
)

求每门课程的学生人数

SELECT CId,COUNT(*) AS '学生人数' FROM sc
GROUP BY CId

成绩不重复，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 s.*, MAX(sc.score) FROM student s,sc,course c,teacher t
WHERE s.SId=sc.SId AND sc.CId=c.CId AND t.TId=c.TId AND Tname='张三'

成绩有重复的情况下，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 student.*, sc.score, sc.cid from student, teacher, course,sc 
where teacher.tid = course.tid
and sc.sid = student.sid
and sc.cid = course.cid
and teacher.tname = "张三"
and sc.score = (
    select Max(sc.score) 
    from sc,student, teacher, course
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
)

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  a.cid, a.sid,  a.score from sc as a
inner join 
sc as b
on a.sid = b.sid
and a.cid != b.cid
and a.score = b.score
group by cid, sid;

查询每门功课成绩最好的前两名

select a.sid,a.cid,a.score from sc as a 
left join sc as b 
on a.cid = b.cid and a.score<b.score
group by a.cid, a.sid
having count(b.cid)<2
order by a.cid;

统计每门课程的学生选修人数（超过 5 人的课程才统计）

SELECT CId,COUNT(*) FROM sc GROUP BY CId HAVING COUNT(*)>5

检索至少选修两门课程的学生学号

SELECT SId FROM sc GROUP BY SId HAVING COUNT(*)>=2

查询选修了全部课程的学生信息

语句转换：不存在这样的学生——所有课程与他选修的课程不存在相等的情况。

SELECT * FROM student WHERE NOT EXISTS
(
SELECT * FROM course WHERE NOT EXISTS
(
SELECT * FROM sc WHERE sc.SId=student.SId AND course.CId=sc.CId
)
)

查询各学生的年龄，只按年份来算

select sname, year(now())-year(sage) as age from student

**OVER！**好像大概使用了四天完成的吧。所有都是自己在navicat中测试过的，如果有不对的还望读者帮忙指正，不会的一起交流！