--数据表
CREATE TABLE `tem` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`str` char(1) DEFAULT NULL,
PRIMARY KEY (`id`)
) ;


INSERT INTO `test`.`tem`(`id`, `str`) VALUES (1, 'A');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (2, 'B');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (3, 'A');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (4, 'C');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (5, 'A');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (6, 'C');
INSERT INTO `test`.`tem`(`id`, `str`) VALUES (7, 'B');

SELECT
    @num := @num+1 num,
    id,
    str
FROM
    tem, (SELECT @str := '', @num := 0) t1
ORDER BY
    str, id;

--变量方式
SELECT
    @num := IF(@str = str, @num + 1, 1) num,
    id,
    @str := str str
FROM
    tem, (SELECT @str := '', @num := 0) t1
ORDER BY
    str, id;

--子查询方式【取分组中前N行(排名前几名)】
select * from testGroup as a
where a.ID in (select  ID from testGroup b where a.UserID = b.UserID order by b.OrderID limit 2)
--或者
select * from testGroup a
where not exists (select 1 from testGroup b where a.UserID = b.UserID and a.OrderID > b.OrderID 
having count(1) >= 2)
--或者
select * from testGroup a
where (select count(1) from testGroup b where a.UserID = b.UserID and a.ID >= b.ID) <= 2
--没有唯一标识的表，可以用checksum来标识每行(MSSQL?)
select * from testGroup as a
where checksum(*) in (select top 2 checksum(*) from testGroup b where a.UserID = b.UserID order by b.OrderID)

create table test1_1(id int auto_increment primary key,`subject` char(20),score int);
insert into test1_1 values(null,'语文',99),(null,'语文',98),(null,'语文',97);
insert into test1_1 values(null,'数学',89),(null,'数学',88),(null,'数学',87);
insert into test1_1 values(null,'英语',79),(null,'英语',78),(null,'英语',77);

-- 根据成绩，求出每个科目的前2名

select * from test1_1;

select * from test1_1 t1
where (select count(1) from test1_1 t2 where t1.subject=t2.subject and t2.score>=t1.score ) <=2;

--变量会最后再计算，所以是先排序好之后，才会开始计算@num

select id,@num:=IF(@STR=STR,@num,@num+1) rn,@str:=str str
from tem t1,(select @str:='',@num:=0) t2 
order by str 

--错误的方式
select id,str,
case when @str=str then @num
when @str:=str then @num:=@num+1 end as 'rn'
from tem t1,(select @str:='',@num:=1) t2 
order by str

--正确的方式
select id,str,
case when @str=str then @num
when @str:=str then @num:=@num+1 
else @num:=@num+1 end as 'rn'
from tem t1,(select @str:='',@num:=0) t2 
order by str

select 
*,
--score,
@a:=@a +(@pre<>(@pre:=score)) as rank
from
score,
(select @a:=0,@pre:=-1) t
order by 
score desc;