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oracle分析函数(二)
作者:武汉SEO闵涛  文章来源:敏韬网  点击数1998  更新时间:2009/4/22 22:03:16  文章录入:mintao  责任编辑:mintao
2. rank函数的介绍

介绍完rollup和cube函数的使用,下面我们来看看rank系列函数的使用方法.

问题2.我想查出这几个月份中各个地区的总话费的排名.


  Quote:
为了将rank,dense_rank,row_number函数的差别显示出来,我们对已有的基础数据做一些修改,将5763的数据改成与5761的数据相同.
  1  update t t1 set local_fare = (
  2    select local_fare from t t2
  3     where t1.bill_month = t2.bill_month
  4     and t1.net_type = t2.net_type
  5     and t2.area_code = ''''5761''''
  6* ) where area_code = ''''5763''''
07:19:18 SQL> /

8 rows updated.

Elapsed: 00:00:00.01

我们先使用rank函数来计算各个地区的话费排名.
07:34:19 SQL> select area_code,sum(local_fare) local_fare,
07:35:25   2    rank() over (order by sum(local_fare) desc) fare_rank
07:35:44   3  from t
07:35:45   4  group by area_codee
07:35:50   5
07:35:52 SQL> select area_code,sum(local_fare) local_fare,
07:36:02   2    rank() over (order by sum(local_fare) desc) fare_rank
07:36:20   3  from t
07:36:21   4  group by area_code
07:36:25   5  /

AREA_CODE      LOCAL_FARE  FARE_RANK
---------- -------------- ----------
5765            104548.72          1
5761             54225.41          2
5763             54225.41          2
5764             53156.77          4
5762             52039.62          5

Elapsed: 00:00:00.01

我们可以看到红色标注的地方出现了,跳位,排名3没有出现
下面我们再看看dense_rank查询的结果.


07:36:26 SQL> select area_code,sum(local_fare) local_fare,
07:39:16   2    dense_rank() over (order by sum(local_fare) desc ) fare_rank
07:39:39   3  from t
07:39:42   4  group by area_code
07:39:46   5  /

AREA_CODE      LOCAL_FARE  FARE_RANK
---------- -------------- ----------
5765            104548.72          1
5761             54225.41          2
5763             54225.41          2
5764             53156.77          3  这是这里出现了第三名
5762             52039.62          4

Elapsed: 00:00:00.00


在这个例子中,出现了一个第三名,这就是rank和dense_rank的差别,
rank如果出现两个相同的数据,那么后面的数据就会直接跳过这个排名,而dense_rank则不会,
差别更大的是,row_number哪怕是两个数据完全相同,排名也会不一样,这个特性在我们想找出对应没个条件的唯一记录的时候又很大用处


  1  select area_code,sum(local_fare) local_fare,
  2     row_number() over (order by sum(local_fare) desc ) fare_rank
  3  from t
  4* group by area_code
07:44:50 SQL> /

AREA_CODE      LOCAL_FARE  FARE_RANK
---------- -------------- ----------
5765            104548.72          1
5761             54225.41          2
5763             54225.41          3
5764             53156.77          4
5762             52039.62          5

在row_nubmer函数中,我们发现,哪怕sum(local_fare)完全相同,我们还是得到了不一样排名,我们可以利用这个特性剔除数据库中的重复记录.

这个帖子中的几个例子是为了说明这三个函数的基本用法的. 下个帖子我们将详细介绍他们的一些用法.




2. rank函数的介绍

a. 取出数据库中最后入网的n个用户
select user_id,tele_num,user_name,user_status,create_date
from (
   select user_id,tele_num,user_name,user_status,create_date,
      rank() over (order by create_date desc) add_rank
   from user_info
)
where add_rank <= :n;

b.根据object_name删除数据库中的重复记录
create table t as select obj#,name from sys.obj$;
再insert into t1 select * from t1 数次.
delete from t1 where rowid in (
   select row_id from (
      select rowid row_id,row_number() over (partition by obj# order by rowid ) rn
   ) where rn <> 1
);

c. 取出各地区的话费收入在各个月份排名.
SQL> select bill_month,area_code,sum(local_fare) local_fare,
  2     rank() over (partition by bill_month order by sum(local_fare) desc) area_rank
  3  from t
  4  group by bill_month,area_code
  5  /

BILL_MONTH      AREA_CODE           LOCAL_FARE  AREA_RANK
--------------- --------------- -------------- ----------
200405          5765                  25057.74          1
200405          5761                  13060.43          2
200405          5763                  13060.43          2
200405          5762                  12643.79          4
200405          5764                  12487.79          5
200406          5765                  26058.46          1
200406          5761                  13318.93          2
200406          5763                  13318.93          2
200406          5764                  13295.19          4
200406          5762                  12795.06          5
200407          5765                  26301.88          1
200407          5761                  13710.27          2
200407          5763                  13710.27          2
200407          5764              

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