Chinaunix首页 | 论坛 | 博客
  • 博客访问: 632641
  • 博文数量: 825
  • 博客积分: 5000
  • 博客等级: 大校
  • 技术积分: 4980
  • 用 户 组: 普通用户
  • 注册时间: 2008-10-27 14:19
文章分类

全部博文(825)

文章存档

2011年(1)

2008年(824)

我的朋友

分类:

2008-10-27 14:28:36


  1. 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. 三个函数的基本用法
  
  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         13444.09     4
  200407     5762         13224.30     5
  200408     5765         27130.64     1
  200408     5761         14135.78     2
  200408     5763         14135.78     2
  200408     5764         13929.69     4
  200408     5762         13376.47     5
  
  20 rows selected.
  SQL>
  
  3. lag和lead函数介绍
  
  取出每个月的上个月和下个月的话费总额
  
  1 select area_code,bill_month, local_fare cur_local_fare,
  2   lag(local_fare,2,0) over (partition by area_code order by bill_month ) pre_local_fare,
  3   lag(local_fare,1,0) over (partition by area_code order by bill_month ) last_local_fare,
  4   lead(local_fare,1,0) over (partition by area_code order by bill_month ) next_local_fare,
  5   lead(local_fare,2,0) over (partition by area_code order by bill_month ) post_local_fare
  6 from (
  7   select area_code,bill_month,sum(local_fare) local_fare
  8   from t
  9   group by area_code,bill_month
  10* )
  SQL> /
  AREA_CODE BILL_MONTH CUR_LOCAL_FARE PRE_LOCAL_FARE LAST_LOCAL_FARE NEXT_LOCAL_FARE POST_LOCAL_FARE
  --------- ---------- -------------- -------------- --------------- --------------- ---------------
  5761   200405     13060.433       0        0    13318.93    13710.265
  5761   200406      13318.93       0    13060.433    13710.265    14135.781
  5761   200407     13710.265   13060.433    13318.93    14135.781        0
  5761   200408     14135.781    13318.93    13710.265        0        0
  5762   200405     12643.791       0        0    12795.06    13224.297
  5762   200406      12795.06       0    12643.791    13224.297    13376.468
  5762   200407     13224.297   12643.791    12795.06    13376.468        0
  5762   200408     13376.468    12795.06    13224.297        0        0
  5763   200405     13060.433       0        0    13318.93    13710.265
  5763   200406      13318.93       0    13060.433    13710.265    14135.781
  5763   200407     13710.265   13060.433    13318.93    14135.781        0
  5763   200408     14135.781    13318.93    13710.265        0
【责编:admin】

--------------------next---------------------

阅读(385) | 评论(0) | 转发(0) |
给主人留下些什么吧!~~