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  • mysql 入门练习(2)

    //重建数据库

    101, create database testdatabase;use database testdatabase;
    
    102, create table tt1(id int, name varchar(20),age int,sex boolean);
    
    103, show tables;desc tt1;

    //插入

    104, insert into tt1 values(1,"zhang",25,0);
    
    105, insert into tt1 values(2,"wang",25,1);
    
    106, select * from tt1;
    
    107, insert into tt1(id,name,age,sex) values(3,"li",28,1);
    
    108, insert into tt1(id,name,sex,age) values(4,"sun",0,22);
    
    109, insert into tt1(id,name,sex) values(5,"zhao",30,1);
    
    110, insert into tt1(id,age,name) values(6,"he",47,0);
    
    111, insert into tt1(id,age,name) values(7,"chen",22,1),(7,"zhang",22,1),(7,"xie",32,1);
    
    112, select * from tt1;

    //修改

    113, update tt1 set id=10,name="new",age=100,sex=0 where id=1; select * from tt1;
    
    114, update tt1 set id=11,name="new" where id=2,age=25; select *from tt1;
    
    115, update tt1 set id=12,sex=1 where id=7; select * from tt1;
    
    116, update tt1 set sex=1 where id>3;
    
    117, update tt1 set sex=0 where id<4;

    //删除

    118, delete from tt1 where id=1;select * from tt1;
    
    119, delete from tt1 where id=12;select * from tt1;

    //查询

    120, alter table tt1 add address varchar(30);
    
    121, update tt1 set address="Beijing" where sex=1;
    
    122, update tt1 set address="Shanghai" where sex=0;

    //简单查询

    123, select id from tt1;
    
    124, select id,name from tt1;
    
    125, select id,name,address from tt1;

    //条件查询

    126, select id,name,address from tt1 where address="Beijing";
    
    127, select * from tt1 where id in(2,3,4);
    
    128, select * from tt1 where id not in(2,3,4);
    
    129, select * from tt1 where id between 2 and 5;
    
    130, select * from tt1 where id not between 2 and 5;
    
    131, select * from tt1 where address like "beijing";
    
    132, select * from tt1 where address like "bei";
    
    133, select * from tt1 where address like "bei%";
    
    134, select * from tt1 where address not like "bei%";
    
    135, select * from tt1 where address is null;
    
    136, select * from tt1 where address is not null;
    
    137, select * from tt1 where age<20 and sex=1;
    
    138, select * from tt1 where sex=0 or age>30;

    //查询结果不重复

    139, select distinct address from tt1; 


    //查询结果排序

    140, select * from tt1 order by age;
    
    141, select * from tt1 order by age asc;
    
    142, select * from tt1 order by age desc;

    //分组查询

    143, select * from tt1 group by sex;/

    /单独使用group by 只会选择每个分组的一条记录

    //group by 与 group_concat结合使用

    144, select group_concat(name),sex from tt1 group by sex;
    
    145, select group_concat(name),group_concat(age),sex from tt1 group by sex;
    
    146, select group_concat(name,age),sex from tt1 group by sex;

    //group by与集合函数结合使用

    147, select sex,count(sex) from tt1 group by sex;
    
    148, select sex,count(sex) from tt1 group by sex having sex>2;//having用于对分组后的结果加限制条件
    
    149, select sex,count(sex) from tt1 group by sex having count(sex)>2;

    //with rollup

    150, select sex,count(sex) from tt1 group by sex with rollup

    ;//在显示记录的最后一行加一条,记录是上面所有记录的和,通常用于合计数量
    //limit显示限制

    151, select * from tt1;
    
    152, select * from tt1 limit 2;
    
    153, select * from tt1 limit 3;
    
    154, select * from tt1 limit 0,2;
    
    155, select * from tt1 limit 1,2;

    //使用集合函数查询

    //为了更好的理解本问题,新建一个表

    156, create table grade(id int,name varchar(10),subject varchar(10),score int,sex boolean);
    
    157, insert into grade values(1,"wang","math",100,1),(1,"wang","english",96,1),(1,"wang","physics",90,1);
    
    insert into grade values(2,"li","math",96,1),(2,"li","english",85,1),(2,"li","physics",99,1);
    
    insert into grade values(3,"sun","math",85,0),(3,"sun","english",98,0),(3,"sun","physics",80,0);
    
    158, select * from grade;
    
    159, select count(*) from grade;
    
    160, select id,name,sum(score) from grade where id=1;
    
    161, select id,name,sun(score) from grade group by id;
    
    162, select id,name,sum(score) from grade group by id order by sum(score) desc;
    
    163, select id,name,avg(score) from grade where id=2;
    
    164, select id,name,avg(score),sum(score) from grade where id =3;
    
    165, select id,name,avg(score) from grade group by id;
    
    166, select subject,avg(score) from grade group by subject;* from 
    
    167, select subject,avg(score) from grade group by subject order by avg(score);
    
    168, select name,max(score) from grade where subject="math";
    
    169, select name,max(score) from grade where subject="english";

    //连接查询

    //内连接

    170, create table stu(id int,name varchar(10),age int,sex boolean);
    
    171, insert into stu values(1,"wang",25,1),(2,"li",23,1),(3,"sun",23,0),(4,"zhou",27,1),(5,"zhang",22,0);
    
    172, select id,name,age,sex,score from stu,grade where stu.id=grade.id;
    
    173, select stu.id,stu.name,stu.age,stu.sex,score from stu,grade where stu.id=grade.id;
    
    174, select stu.id,stu.name,stu.age,stu.sex,score from stu,grade where stu.id=grade.id and score>90;

     

    //外连接

    175, select stu.id,stu.name,stu.age,stu.sex,score from stu left join grade on stu.id=grade.id;
    
    176, select stu.id,stu.name,stu.age,stu,sex,score from stu right join grade on stu.id=grade.id;
    
    177, insert into grade values(4,"hu","math",99,1),(5,"liu","english",97,0);
    
    178, select stu.id,stu.name,stu.age,stu.sex,subject,score from stu left join grade on stu.id=grade.id;
    
    179, select stu.id,stu.name,stu.age,stu.sex,subject,score from stu right join grade on stu.id=grade.id;

    //子查询

    180, select * from stu where id in (select *from grade);
    
    181, select * from grade where id in (select * from stu);
    
    182, create table scholarship(level int,money int,score int);
    
    183, insert into scholarship values(1,10000,90),(2,5000,85),(3,3000,80);
    
    184, select id,name,score from grade where score>=(select score from scholarship where level=1);
    
    185, select id,name,score from grade where score>=(select min(score) from scholarship);

    //exists子查询

    186, select * from stu where exists (select name from grade where id=1);
    
    187, select * from grade where score>any(select score from scholarship);
    
    188, select * from grade where score>all(select score from scholarship);

    //合并查询结果

    189, select name from stu union select name from grade;
    
    190, select name from stu union all select name from grade;

    //别名

    191, select * from scholarship s where s.level=1;
    
    192, select * from scholarship s where s.money=5000;
    
    193, select s.name from stu s,grade g where s.name=g.name;
    
    194, select subject as su,score as sc from grade;
    
    195, select * from stu where name regexp "^w";
    
    196, select * from stu where name regexp "g$";
    
    197, select * from stu where name regexp "^w.g$";
    198, select * from stu where name regexp "^w..g$";
    199, select * from stu where name regexp "[aeo]";
    200, select * from stu where name regexp "[a-g]";
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  • 原文地址:https://www.cnblogs.com/Koi504330/p/11901577.html
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