Mysql对表中数据查询的操作 DQL

准备数据，倒入sql文件

运行sql文件得到四张表

select * from 表名 * 代表全部

1、AS子句作为别名

select studentname as "姓名" , sex as "性别",gradeid as "班级ID" from student;

多表查询给表起别名 简化代码

2、distinct 去重

DISTINCT关键字
    作用：
        去掉SELECT查询返回的记录结果中重复的记录（所有返回列的值都相同），只返回一条

    语法：

多条语句去重

select distinct ID,NAME from student;

以 ID和NAME 为单位

例如：如下不算重复
1 aa
1 bb

3、where条件语句

4、LIKE模糊查询

查询姓李的同学记录
% 所有字符
select * from student where StudentName like "李%";


"_" 代表一个字符

查询姓李X(名只有1个)的同学记录 
select * from student where StudentName like "李_";

查询名字中包含文的记录
select * from student where StudentName like "%文%";

查询以 姓名 明 结尾的纪录
select * from student where StudentName like "%明"

查询所有姓李的学生所有成绩
select s.StudentNo as "'李'同学学号",r.stuResult as "'李'同学成绩" from student as s ,result as r
where s.StudentNo=r.StudentNo AND s.StudentName like "李%";

查询 Java的考试成绩

select s.subjectName as java,r.stuResult as "java的成绩" from subject as s ,result as r
where s.SubjectName="java" and s.SubjectNo=r.SubjectNo;

查询考试JAVA的学生的学号和考试成绩
select r.StudentNo as "学号",r.stuResult AS "考试成绩" from result as r ,subject as s
where r.subjectNo=s.subjectNo and s.subjectName="java";

查询考试Java的学生学号，姓名，考试成绩

select r.StudentNo as "学号",st.StudentName as "姓名",r.stuResult AS "考试成绩"
from result as r ,subject as s,student as st where
r.subjectNo=s.subjectNo AND
r.StudentNo=st.StudentNo and
s.subjectName="java";

5 查询Student表中年龄为12或性别为“女”的同学记录。
CURDATE() 函数返回当前的日期。

select * from student where TIMESTAMPDIFF(YEAR,BornDate,CURDATE())=12 or Sex=0;

year(now())-year(BronDate)

TIMESTAMPDIFF函数，有参数设置，

可以精确到天（DAY）、小时（HOUR），分钟（MINUTE）和秒（SECOND），

。对于比较的两个时间，时间小的放在前面，时间大的放在后面。

5、联合查询

SELECT 字段列表
    FROM 表1 INNER|LEFT|RIGHT JOIN 表2
    ON 表1.字段 = 表2.字段;

要求：从subject和grade数据表查询课程名称和所属年级名称

select s.subjectName as "课程名称" ,g.gradeName as "年级名称" from subject as s INNER JOIN grade as g
on s.gradeid=g.gradeID;

JION左边的是左表 JOIN JOIN右边的是右表

6、order by排序

查询所有考试结果，并按成绩由高到低排列
ORDRY BY 字段
select * from result ORDER BY stuResult desc;

根据成绩升序排序，当成绩相同时根据学号降序排序
主要排序和次要排序
select * from result ORDER BY stuResult asc ,StudentNo DESC ;

7、LIMIT 分页查询

m对应索引，n是步长

select * from result limit 1,5;

查询起始为学号为1002 共5条记录

8、统计函数

count计数、sum求和、avg平均值、max最大值、最小值

SELECT count(*) from student;

count可以不放字段名，其他的都要放字段

9、group by 分组

按照不同的课程分组，分别算出其平均分、最高分和最低分，对于
平均分低于60分的不予显示

select s.subjectName as "课程名" ,avg(r.stuResult) as "平均分",MAX(r.stuResult) "最高分",MIN(r.stuResult) "最低分"
from subject s, result r where s.SubjectNo=r.SubjectNo  GROUP BY r.SubjectNo HAVING avg(r.stuResult)>60;

10、having


普通等值写法

select subjectName,avg(stuResult),max(stuResult),min(stuResult) from subject s,result r
WHERE s.subjectNo=r.subjectNO
GROUP BY r.subjectno
HAVING avg(stuResult)>60;

连表方法

select subjectname,avg(stuResult),max(stuResult),min(stuResult) from subject s
INNER JOIN result r
on r.subjectno=s.subjectno
GROUP BY r.subjectno
having avg(stuResult)>60;

11、子查询

查询大二的上了什么课？

select subjectName from subject where gradeid =
(select gradeID from grade where gradeName="大二");

查询课程为java且分数不小于80分的学生的学号和姓名？
gradeid是桥梁

子查询方法：
SELECT StudentNo,StudentName from student where StudentNo in
(select StudentNo from result where SubjectNo in
(select subjectNo FROM subject where subjectName="java") and stuResult>=80
);

连表查询方法：
select st.StudentNo,StudentName from student as st
INNER JOIN
result as r on st.StudentNo=r.StudentNo
INNER JOIN subject as s ON
r.SubjectNo=s.subjectNo where s.subjectName="java" and r.stuResult>=80;

等值连接
select s.studentno,studentname from student s,subject su,result r
WHERE r.studentno=s.studentno
and su.subjectno=r.subjectno
and subjectname='java'
and stuResult>80;

sql语句注意事项

(1)MYSQL 查询语句不区分大小写

例如：

select * from table where name = "aa"

select *from table where name = "AA"

这两条的查询结果是一样的

(2)sql count(*) 和 count(col)、count(1)区别

row行，col列

count(*)和count(列)根本就是不等价的，count(*)是针对于全表的，而count(列)是针对于某一列的，如果此列值为空的话，count(列)是不会统计这一行的。

也就是说count(列)会用到索引，并且会过滤掉该列为null的那行。

而count(*)是全表扫描，并且只要某一条数据有一列不为null，那就会统计到这条数据。

 
count(1)可以理解为表里有一列数据都为1的列。

count(*)，执行时会把星号翻译成字段的具体名字，效果也是一样的，不过多了一个翻译的动作，比固定值的方式效率稍微低一些。

group_concat 和 group by 是一起使用的

综合练习

course sc student teacher

course:Cid,Cname,Tid
sc:Id,Sid,Cid,score
student:Sid,Sname,Sage,Sex
teacher:Tid,Tname


1查询每门课程被选修的人数
select Cid,count(Sid) "人数" from sc GROUP BY Cid;
select Cid,count(*) "人数" from sc GROUP BY Cid;

2 查询只选修了4门课程的学生学号，和姓名
SELECT s.Sid,s.Sname from student s INNER JOIN sc on sc.Sid=s.Sid
GROUP BY sc.Sid HAVING count(sc.Cid)=4;

select s.sid,sname from student s
INNER JOIN sc
on s.Sid=sc.Sid
GROUP BY s.Sid
HAVING count(*)=4;

3 查询姓张的学生名单
select Sname from student where Sname like "李%";


4 查询同名同姓的学生名单，显示姓名，同名的人数
select Sname,count(Sid) "人数" from student GROUP BY Sname HAVING count(Sid)>=2;

select sname,count(*) from student 
GROUP BY sname
having count(*)>=2;

5 查询每门课程的平均成绩，按照平均成绩升序，平均成绩相同时按课程号降序；

select Cname "课程名",avg(score) "平均成绩" from course c INNER JOIN sc on
c.Cid=sc.Cid GROUP BY c.Cid ORDER BY avg(score),c.Cid desc;

select cid,avg(score) FROM sc 
GROUP BY cid
ORDER BY avg(score) asc,cid desc;

6 查询平均成绩大于80分的学生的学号，姓名和平均成绩

select s.Sid "学号",Sname "姓名",avg(score) "平均成绩" from student s INNER JOIN sc ON
s.Sid=sc.Sid GROUP BY s.Sid HAVING avg(score)>80;

select s.sid,sname,avg(score) 'avg' FROM sc
INNER JOIN student s
on s.sid=sc.sid
GROUP BY s.sid
HAVING avg>80;

7 查询  所有  的课程分数都低于80分的学生的姓名 ???????已解决
是所有的！！！
select Sname "姓名" from student s INNER JOIN sc ON
s.Sid = sc.Sid GROUP BY s.Sid HAVING MAX(score)<80;


select sname from sc 
INNER JOIN student s
on s.sid=sc.Sid
GROUP BY s.Sid
HAVING max(score)<80;



8 查询所有同学的选课情况，显示学生的姓名和选修的课程名


select Sname "姓名",Cname "课程名" from student s INNER JOIN sc ON
s.Sid=sc.Sid INNER JOIN course c ON c.Cid=sc.Cid ORDER BY s.Sid;

下面这种更好：
select sname,group_concat(cname) from student s
INNER JOIN sc 
on s.sid=sc.sid
INNER JOIN course c
on c.cid=sc.cid
GROUP BY s.sid

 group_concat 和 group by 是一起使用的

9 查询学习Java的学生人数。

select count(Sid) "学习java人数" from sc INNER JOIN course on 
course.Cid=sc.Cid WHERE course.Cname="java";


select count(*) from sc 
INNER JOIN course c
on sc.cid=c.cid
WHERE cname='java';
10查询sc表中的最高分的学生学号和课程号。（子查询或者排序）


select s.Sid "最高分学号",sc.Cid "课程号" from student s INNER JOIN sc ON
s.Sid=sc.Sid WHERE score = (SELECT MAX(score) from sc);


select sid,cid,score from sc 
ORDER BY score DESC
limit 1;

11 查询成绩在80-90之间的学生学号
更正去重
SELECT DISTINCT Sid from sc WHERE score BETWEEN 80 AND 90;




12 查询成绩在80-90之间的学生的姓名
更正去重
SELECT DISTINCT Sname "学生姓名" from student INNER JOIN sc on student.Sid=sc.Sid
WHERE score BETWEEN 80 AND 90;


SELECT DISTINCT sname from student s
INNER JOIN sc
on s.sid=sc.sid
WHERE score BETWEEN 80 and 90;


13 查询每个学生的平均成绩

select Sname "学生姓名",avg(score) "平均成绩" from student s INNER JOIN sc ON
s.Sid=sc.Sid  GROUP BY sc.Sid;


14 查询所有任课老师的姓名和课程名

select Tname "老师姓名",Cname "课程名" from teacher t INNER JOIN course c
on t.Tid=c.Tid;

下面的方法更好：
select tname,group_concat(cname) from teacher t
INNER JOIN course c
on c.tid=t.tid
GROUP BY t.tid;

15 查询最高分的学生学号，学生姓名，考试成绩 

错误：
select s.Sid "学号",Sname "学生姓名",score "考试成绩" from student s INNER JOIN sc
on s.Sid=sc.Sid where score = (SELECT MAX(score) from sc);


更正：

select sname,s.sid,score from sc 
INNER JOIN student s
on s.sid=sc.sid
WHERE s.sid=(
select sid from sc 
WHERE score=(select max(score) from sc)
);


select sname,s.sid,GROUP_CONCAT(score) from sc 
INNER JOIN student s
on s.sid=sc.sid
WHERE s.sid=(
select sid from sc 
WHERE score=(select max(score) from sc)
)
GROUP BY s.sid;

练习题2，重点

course:cid,cname,tid
sc:sid,cid,score
student:sid,sname,age,sex
teacher:tid,tname

select * from course;
select * from sc;
select * from student;
select * from teacher;

1．查询平均成绩大于60分的同学的学号和平均成绩????  要分组
where只能用于存在的列 错误表达： WHERE AVG(score)>60;

select s.sid,avg(score) from student s INNER JOIN sc
on s.sid=sc.sid HAVING group by s.sid avg(score)>60;

2．查询所有同学的学号、姓名、选课数、总成绩

select s.sid "学号",sname "姓名",count(cid) "选课数",sum(score)"总成绩"
from student s INNER JOIN sc on s.sid=sc.sid GROUP BY sc.sid

3．    查询姓“李”的老师的个数

select count(*)"姓李老师个数" from teacher where tname LIKE "李%";

5.查询学过“李纯”老师课的同学的学号、姓名

select s.sid"学号",sname"姓名" from student s INNER JOIN sc ON
s.sid=sc.sid INNER JOIN course as c on c.cid=sc.cid INNER JOIN
teacher t on c.tid = t.tid WHERE tname = "李纯"; 


6.查询没学过“李雷”老师课的同学的学号、姓名?????????---已解决

select sid,sname from student where sid not in (
select sid from sc where cid in(
select cid from course where tid =(
select tid from teacher where tname="李雷")));

7.查询“001”课程比“002”课程成绩高的所有学生的学号、姓名?????? --已解决---
course:cid,cname,tid
sc:sid,cid,score
student:sid,sname,age,sex
teacher:tid,tname

法一：

select sid,sname from student where sid in(
SELECT t1.sid FROM sc t1, sc t2
WHERE
t1.sid=t2.sid and
t1.cid = '001' AND
t2.cid = '002' AND
t1.score > t2.score);

法二：
select s.sid,sname from student s
INNER JOIN sc a on s.sid=a.sid INNER JOIN sc b
on a.sid=b.sid
where
a.cid = '001' AND
b.cid = '002' AND
a.score > b.score;

8. 查询学过“001”并且也学过编号“003”课程的同学的学号、姓名

select s.sid"学号",sname"姓名" from student s INNER JOIN sc a ON
s.sid =a.sid INNER JOIN sc b on a.sid=b.sid
WHERE a.cid="001" and b.cid="003" GROUP BY b.sid;

9.查询各科成绩最高分和最低分：以如下形式显示：课程ID，课程名，最高分，最低分

select c.cid"课程ID",cname"课程名",max(score)"最高分",min(score)"最低分" FROM
course c INNER JOIN sc on c.cid=sc.cid GROUP BY sc.cid;


10. 查询没有学全所有课的同学的学号、姓名

select s.sid"学号",sname"姓名" from student s INNER JOIN sc ON
s.sid =sc.sid GROUP BY sc.sid HAVING count(*)<(select count(*) from course);

11.查询学过“李雷”老师所教的所有课的同学的学号、姓名？？？----已解决
course:cid,cname,tid
sc:sid,cid,score
student:sid,sname,age,sex
teacher:tid,tname

select sid"学号",sname"姓名" from student where sid in (
select sid from sc where cid in (
SELECT cid from course where tid=
(select tid from teacher where tname="李雷")));


12.    查询至少学过学号为“1001”同学所学的一门课的其他同学学号和姓名

方法一：

要排除学号为"1001"的学号

select DISTINCT s.sid"学号",sname"姓名" from student s INNER JOIN sc ON
s.sid=sc.sid where cid in
(select cid from sc where sid = "1001") and s.sid<>"1001";


方法三：
select  sid"学号",sname"姓名" from student WHERE sid in
(select DISTINCT sid from sc where cid in
(select cid from sc where sid = "1001")and sid<>"1001");

13.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排序???????---已解决


select cid,round(avg(score),2)t1,(sum(
case when score BETWEEN 60 and 100 then 
1
ELSE
0
end)/count(*))*100  t2
from sc GROUP BY cid ORDER BY t1,t2 desc;

select cid,round(avg(score),2)t1,CONCAT(round((sum(
case when score BETWEEN 60 and 100 then 
1
ELSE
0
end)/count(*))*100),'%')  t2
from sc GROUP BY cid ORDER BY t1,t2 desc;

14.删除学习“李纯”老师课的SC表记录
DELETE from sc where cid in (
SELECT cid from course where tid=
(select tid from teacher where tname="李纯"));

练习题三

select * from course;
select * from sc;
select * from student;
select * from teacher;
course:cid,cname,tid
sc:sid,cid,score
student:sid,sname,age,sex
teacher:tid,tname

1 查询任何一门课程成绩在70分以上的姓名、课程名和分数
select sname,GROUP_CONCAT(cname),GROUP_CONCAT(score) from student s INNER JOIN sc ON
s.sid=sc.sid INNER JOIN course c on c.cid=sc.cid GROUP BY sc.sid  HAVING min(sc.score)>70;

2 查询所有不及格的课程，并按课程号从大到小罗列课程ID和学生ID

select cid,GROUP_CONCAT(s.sid) from sc INNER JOIN student s on sc.sid=s.sid
where score<60 GROUP BY cid ORDER BY cid desc;

3 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,sname from student s INNER JOIN sc on s.sid=sc.sid
WHERE cid='003' and score>80;

4 求选全了课程的学生人数
select count(a.sid) from 
(select sid from sc 
 GROUP BY sid HAVING COUNT(*)=
(select count(*) from course)) a;

5 查询选修李 老师所授课程的学生中，成绩最高的学生姓名及其成绩

select sname,max(score) from student s ,teacher t,course c,sc
where t.tid=c.tid and 
tname like "李%" AND
c.cid=sc.cid AND
sc.sid=s.sid 
GROUP BY tname;


6 查询各个课程及相应的选修人数
select c.cid,count(*) from course c INNER JOIN sc ON
c.cid=sc.cid GROUP BY sc.cid;

7 查询不同课程成绩相同的学生的学号、课程号、学生成绩?????
select s1.sid,s2.sid,sc1.cid,sc1.score,sc2.score from student s1,student s2,sc sc1,sc sc2 
where s1.sid!=s2.sid and s1.sid=sc1.sid and s2.sid=sc2.sid and sc1.cid!=sc2.cid and sc1.score=sc2.score;

8 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;

9 查询全部学生都选修的课程的课程号和课程名
哪一个课程被所有学生多选？

select c.cid,cname from course c INNER JOIN sc ON
c.cid=sc.cid  GROUP BY sc.cid HAVING count(*)=(select count(*) from student);

4 求选全了课程的学生人数????
select count(*) from 
(select sid from sc GROUP BY sid HAVING
count(*)=(select count(*) from course)) a;

2 查询sc表里 所有 不及格的课程，并按照课程号从大到小罗列课程id和学生id?????
select sid,cid,score from sc where score<60 ORDER BY sid desc,cid desc;

练习四

导入cdsgusTable表

select * from cdsgustable;
id
cname
CtfTp
CtfId
Gender
Birthday
Address
Zip
Mobile
Tel
Fax
EMail
Version

1.    获取每个性别的人数

select gender,count(*) from cdsgustable GROUP BY gender;

2.    查询2012年的数据
select * from cdsgustable where year(version)=2012;

3.    查询2012年12月的数据
select * from cdsgustable where year(version)=2012 and month(version)=12;

4.    查询姓王的人数
select count(*)'姓王的人数' from cdsgustable where cname like '王%'

5.    查询姓李的，并且CTFTP不是OTH的人数

select count(*) from cdsgustable where cname like '李%' and CtfTp<>'OTH';

6.    查询现在年龄大于30岁的人数
select count(a.cname) from 
(select cname from cdsgustable where (year(NOW())-year(Birthday))>30 GROUP BY cname) a;

错误：select count(*) from cdsgustable where (year(NOW())-year(Birthday))>30;


7.    根据出生年月日，转换为星期，现实出生人最多的那个星期几的英语单词
DATE_FORMAT(NOW(),"%W")

select DATE_FORMAT(Birthday,"%W") bir_week from cdsgustable GROUP BY bir_week HAVING count(*)=
(select count(*) from cdsgustable GROUP BY DATE_FORMAT(Birthday,"%W") limit 1);


附加题9：获取每年的男女性别所占的百分比，并保留两位小数（结果如下所示）
        提示：保留几位小数用函数round(值,保留小数位数)

select year(Birthday) year,CONCAT(round((sum(
case when Gender='F' then 
1
ELSE
0
end)/count(Gender))*100,2),'%')'女',CONCAT(round((sum(
case when Gender='M' then 
1
ELSE
0
end)/count(Gender))*100,2),'%')'男'
from cdsgustable GROUP BY year;

select * from course;
select * from sc;
select * from student;
select * from teacher;

course:cid,cname,tid
sc:Id,Sid,Cid,score
student:Sid,Sname,Sage,Sex
teacher:Tid,Tname

1.    查询不同课程成绩相同的学生的学号、课程号、学生成绩
select s1.sid,s1.cid,s2.sid,s2.cid,s1.score from sc s1 INNER JOIN sc s2 on s1.cid<>s2.cid 
where s1.score=s2.score;

2.    统计各科成绩的各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60] 
select sc.cid,cname,
sum(case when  score between 85 and 100 then 1 ELSE 0
end)'[100-85]',
sum(case when  score between 70 and 85 then 1 ELSE 0
end)'[85-70]',
sum(case when  score between 60 and 70 then 1 ELSE 0
end)'[70-60]',
sum(case when  score <60 then 1 ELSE 0
end)'[ <60] ' from sc INNER JOIN course c on c.cid=sc.cid
GROUP BY sc.cid;

3.    查询所有学生的选课情况，显示学生姓名和选修的课程名
select sname,GROUP_CONCAT(cname)from student s INNER JOIN sc ON
s.sid=sc.sid INNER JOIN course c on sc.cid=c.cid GROUP BY sc.sid;

4.    把sc表中李浩然老师教课的成绩都改为此课程的平均成绩 ？？？？

posted on 2019-08-17 21:59 foremost 阅读(676) 评论(0) 编辑收藏举报

Mysql对表中 数据 查询的操作 DQL