文章目录

清空一个表,自增id从1开始
truncate table 表名;

查询
select 列名 from 表名 where 条件 order by 列名 [desc|asc] limit 跳过条数,查多少条

AS 别名
    列名 as 新列名   注意as可以省掉不写
NULL值查询
    select * from table1 where 字段 is null;
组合列
    select concat('No.',id) from stu;
去重复
    select distinct 列名 from 表名;      注意:列名,只能跟一个
排序
    select * from 表名 order by 列名1 asc,列名2 desc;
    加工之后的信息排序
    select concat('No.',id) from stu order by concat('No.',id);
常用函数
    select concat('My','S','QL');        连接
    select lower('PHP');                转小写
    upper()                                转大写
    select replace('www.mysql.com','www','http://www');    替换
    update stu set name = replace( replace(name,'o','0') ,'i' ,'1');  将name字段所有的o换为0,所有的i换为1

    trim()
    ltrim()
    rtrim()
    select abs(-5)    绝对值
    select adddate('2013-02-28',1);     2013-03-01
    select now();

    通配符  "%" 任意   "_"单个
    select * from stu where name like '张_';

    正则  不支持  \w ...
    select * from stu where name regexp '^a[a-z][a-z]$';

    区间操作  [not] between  ... and  ...
        select *  from stu where  age >=18 and age <=19;
        select *  from stu where  age between 18 and 19;
        select *  from stu where  age not between 18 and 19;

    [not] in (值1,值2,...)

    sum    求和
    avg 平均
    count(*)   计数
        括号中可以是*,也可以列名
        如果括号中是列名,则不会统计该列数据为null值的
    max     最大值
        select max(age) from stu;
    min        最小值

    *分组查询
    一条语句得到男生和女生分别是多少人?
    select sex,count(*) from stu group by sex;    #分组后统计
        注意:字段只能是用于分组的字段(group by 后的字段)  或聚合函数

    按科目分组统计平均成绩

    分组带排序
    select sex,avg(age) as avg from stu  group by sex order by avg desc;

    *分组后筛选 having
    按性别分组,求平均年龄大于等于6
    select sex,avg(age) as avg from stu group by sex having avg>=6;
    按性别分组,求平均年龄大于等于6,按平均年龄从大到小排列
    select sex,avg(age) as avg from stu group by sex having avg>=6 order by avg desc;

    where 与having可以同时出现,where在筛选之前就执行了,having是在分组之后才执行

    显示人数超过15人的班级,按班级人数排序
    select grade_id,count(*) from stu group by grade_id having count(*)>15 order by count(*) desc;
    select grade_id,count(*) total from stu group by grade_id having total>15 order by total desc;

    WHERE -> GROUP BY -> HAVING ->ORDER BY

    成绩表
    create table score (
        id int auto_increment primary key,
        stu_id int,
        subject varchar(50) ,
        score int 
    ) engine=myisam default charset=utf8;

    INSERT INTO score (stu_id,subject,score) values (1,'php',65),(1,'mysql',95),(1,'linux',80),(2,'php',50),(2,'mysql',70),(2,'linux',70);

    统计及格总人数和及格人数的平均分
    select subject, count(*),avg(score) from score  where score>=70 group bysubject;
    统计及格总人数和平均分在80以上

    多个学员不低于90分的班级
    select 班级  from  表名  where score >=90  group by  班级  having count(*)>1

    显示:id,姓名,班级名称
    select stu.id,stu.name,grade.name from stu,grade where stu.grade_id=grade.id;
    显示:id,姓名,班级名称,age大于10的
    select stu.id,stu.name,grade.name,age from stu,grade where stu.grade_id=grade.id and age>10;

    select stu.id,stu.name,grade.name,age from stu
    left join grade on grade.id=stu.grade_id
    where age>10;

    左联接
        左表中的数据,全部出来,就算右表没,没有对应的数据
        select stu.id,stu.name,grade.name,age from stu
    left join grade on grade.id=stu.grade_id

    右联接
    select stu.id,stu.name,grade.name,age from stu
    right join grade on grade.id=stu.grade_id

    用右连接,实现和前面一样的效果
    select stu.id,stu.name,grade.name,age from grade
    right join  stu on grade.id=stu.grade_id

    内联  inner可以省掉
    select stu.id,stu.name,grade.name,age from stu
    inner join grade on grade.id=stu.grade_id
    内联与这样结果相同:  select * 表1,表2 where 条件  

    显示所有参加考试的男生的成绩
    select score.*,stu.name from stu join score on score.stu_id=stu.id where sex=1;
    select score.*,stu.name from stu ,score  where stu.id=score.stu_id and sex=1;
    先查男生的id,然后通过id列表,去成绩表中查询
    select id from stu where sex=1; select * from score where stu_id in (1,6,8);
    select * from score where stu_id in (select id from stu where sex=1);

    查询科目及格人数和不及格人数:
    科目 及格人数   不及格人数
    phy            20            1
    mysql            26            2        

    select subject,sum(case when score<60 then 1 else 0 end) as no,
    sum(case when score>=60 then 1 else 0 end) pass 
    from score group by subject;
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