Question

我有以下内容：

id | name  | Book Read        | Date Read
1    Maria   Dogs love cats     12/31/2008
1    Maria   Cats and Dogs      12/31/2007
1    Maria   Cowboys            12/31/2006
2    Tom     Cowboys            12/31/2008
2    Tom     Indians            12/31/2005
2    Tom     Cats and Dogs      12/31/2003
3    Harry   Raining hard       12/31/2005
3    Harry   Cats and Dogs      12/31/2002
3    Harry   Indians            12/31/2001

如果我执行“SELECT * FROM table GROUP BY id”，我会得到：

id | name  | Book Read         | Date Read
1    Maria   Dogs love cats      12/31/2008
2    Tom     Cowboys             12/31/2008
3    Harry   Raining hard        12/31/2005

但我如何为每个人获得前2本书？即：

id | name  | Book Read         | Date Read
1    Maria   Dogs love cats      12/31/2008
1    Maria   Cats and Dogs       12/31/2007
2    Tom     Cowboys             12/31/2008
2    Tom     Indians             12/31/2005
3    Harry   Raining hard        12/31/2005
3    Harry   Cats and Dogs       12/31/2002

Answer 1

假设你有一个自动增量字段，在我的例子中名为i

select * from table as t1
where (select count(*) from table as t2
       where t1.id = t2.id and t2.i < t1.i) <2

这是另一种选择

select tmp.i,tmp.id,tmp.name,tmp.book_read
from (
  select 
    id,i,name,book_read,
    if( @prev <> id, @r:=1,@r:=@r+1 ) as rank,
    @prev := id
  from table as t
  join (select @r:=null,@prev:=0) as rn
  order by t.id,t.i
) as tmp
where tmp.rank <= 2
order by i,id;

分组依据但指定每个分组ID的行数

1 个答案: