hive 全局排序

不分发数据，使用单个reducer

set mapred.reduce.tasks=;

select *
from dw.dw_app
where
dt>='2016-09-01'
and dt <='2016-09-18'
order by stime
limit ;

包多一层，是用order by

select t.* from
(
select *
from dw.dw_app
where
dt>='2016-09-01'
and dt <='2016-09-18'
and app_id=''
and msgtype = 'role.recharge'
) t
order by t.stime
limit ;

把所有具有相同的行最终都在一个reducer分区中，在在一个reducer中排序。 cluster by column=distribute by column+sort by colum

select *
from dw.dw_app
where
dt>='2016-09-01'
and dt <='2016-09-18'
and app_id=''
and msgtype = 'role.recharge'
cluster by dt
limit ;

查询每天前十名充值用户和充值总额

select t3.*
  from (select t2.*
          from (select dt,
                       account_id,
                       sum(recharge_money) as total_money,
                       row_number() over(partition by dt order by sum(recharge_money) desc) rank
                  from (select dt, account_id, recharge_money
                          from dw.dw_app
                         where dt >= '2016-09-01'
                           and dt <= '2016-09-18'
                           and app_id = ''
                           and msgtype = 'role.recharge' 
　　　　　　　　　　　　　　　　cluster by dt, account_id) t
                 group by dt, account_id) t2
         where t2.rank <= ) t3
 order by t3.dt asc, rank asc limit ;