Oracle开发课题之：分析函数3(Top/Bottom N、First/Last、NTile) （转载）

目录
===============================================
1.带空值的排列
2.Top/Bottom N查询
3.First/Last排名查询
4.按层次查询

一、带空值的排列：

在前面《Oracle开发专题之：分析函数2(Rank、Dense_rank、row_number)》一文中，我们已经知道了如何为一批记录进行全排列、分组排列。假如被排列的数据中含有空值呢？

SQL>?select?region_id,?customer_id,
??2?????????sum(customer_sales)?cust_sales,
??3?????????sum(sum(customer_sales))?over(partition?by?region_id)?ran_total,
??4?????????rank()?over(partition?by?region_id
??5??????????????????order?by?sum(customer_sales)?desc)?rank
??6????from?user_order
??7???group?by?region_id,?customer_id;

?REGION_ID?CUSTOMER_ID?CUST_SALES??RAN_TOTAL???????RANK
----------?-----------?----------?----------?----------?
????????10??????????31??????????????????? 6238901??????????1
????????10??????????26????1808949????6238901??????????2
????????10??????????27????1322747????6238901??????????3
????????10??????????30????1216858????6238901??????????4
????????10??????????28?????986964????6238901??????????5
????????10??????????29?????903383????6238901??????????6

我们看到这里有一条记录的CUST_SALES字段值为NULL，但居然排在第一名了！显然这不符合情理。所以我们重新调整完善一下我们的排名策略，看看下面的语句：

SQL>?select?region_id,?customer_id,
??2?????????sum(customer_sales)?cust_total,
??3?????????sum(sum(customer_sales))?over(partition?by?region_id)?reg_total,
??4?????????rank()?over(partition?by?region_id?
????????????????????????order?by?sum(customer_sales)?desc?NULLS?LAST)?rank
??5????????from?user_order
??6???????group?by?region_id,?customer_id;

?REGION_ID?CUSTOMER_ID?CUST_TOTAL??REG_TOTAL???????RANK
----------?-----------?----------?----------?----------
????????10??????????26????1808949??? ?6238901?????????? 1
????????10??????????27????1322747????6238901?????????? 2
????????10??????????30????1216858????6238901?????????? 3
????????10??????????28?????986964???? 6238901?????????? 4
????????10??????????29?????903383???? 6238901?????????? 5
????????10??????????31???????????????????? 6238901???????????6

绿色高亮处，NULLS LAST/FIRST告诉Oracle让空值排名最后或第一。

注意是NULLS，不是NULL。

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二、Top/Bottom N查询：

在日常的工作生产中，我们经常碰到这样的查询：找出排名前5位的订单客户、找出排名前10位的销售人员等等。现在这个对我们来说已经是很简单的问题了。下面我们用一个实际的例子来演示：

【1】找出所有订单总额排名前3的大客户：

SQL>?select?*
SQL>???from?(select?region_id,
SQL>????????????????customer_id,
SQL>????????????????sum(customer_sales)?cust_total,
SQL>????????????????rank()?over(order?by?sum(customer_sales)?desc?NULLS?LAST)?rank
SQL>???????????from?user_order
SQL>??????????group?by?region_id,?customer_id)
SQL>??where?rank?<=?3;

?REGION_ID?CUSTOMER_ID?CUST_TOTAL???????RANK
----------?-----------?----------?----------
?????????9??????????25????2232703??????????1
?????????8??????????17????1944281??????????2
?????????7??????????14????1929774??????????3

SQL>?

【2】找出每个区域订单总额排名前3的大客户：

SQL>?select?*
??2????from?(select?region_id,
??3?????????????????customer_id,
??4?????????????????sum(customer_sales)?cust_total,
??5?????????????????sum(sum(customer_sales))?over(partition?by?region_id)?reg_total,
??6?????????????????rank()?over(partition?by?region_id
??????????????????????????????? order?by?sum(customer_sales)?desc?NULLS?LAST)?rank
??7????????????from?user_order
??8???????????group?by?region_id,?customer_id)
??9???where?rank?<=?3;

?REGION_ID?CUSTOMER_ID?CUST_TOTAL??REG_TOTAL???????RANK
----------?-----------?----------?----------?----------
?????????5???????????4????1878275????5585641??????????1
?????????5???????????2????1224992????5585641??????????2
?????????5???????????5????1169926????5585641??????????3
?????????6???????????6????1788836????6307766??????????1
?????????6???????????9????1208959????6307766??????????2
?????????6??????????10????1196748????6307766??????????3
?????????7??????????14????1929774????6868495??????????1
?????????7??????????13????1310434????6868495??????????2
?????????7??????????15????1255591????6868495??????????3
?????????8??????????17????1944281????6854731??????????1
?????????8??????????20????1413722????6854731??????????2
?????????8??????????18????1253840????6854731??????????3
?????????9??????????25????2232703????6739374??????????1
?????????9??????????23????1224992????6739374??????????2
?????????9??????????24????1224992????6739374??????????2
????????10??????????26????1808949????6238901??????????1
????????10??????????27????1322747????6238901??????????2
????????10??????????30????1216858????6238901??????????3

18?rows?selected.

三、First/Last排名查询：

想象一下下面的情形：找出订单总额最多、最少的客户。按照前面我们学到的知识，这个至少需要2个查询。第一个查询按照订单总额降序排列以期拿到第一名，第二个查询按照订单总额升序排列以期拿到第一名。是不是很烦？因为Rank函数只告诉我们排名的结果，却无法自动替我们从中筛选结果。

幸好Oracle为我们在排列函数之外提供了两个额外的函数：first、last函数，专门用来解决这种问题。还是用实例说话：

SQL>?select?min(customer_id)
??2?????????keep?(dense_rank?first?order?by?sum(customer_sales)?desc)?first,
??3?????????min(customer_id)
??4?????????keep?(dense_rank?last?order?by?sum(customer_sales)?desc)?last
??5????from?user_order
??6???group?by?customer_id;

?????FIRST???????LAST
----------?----------
????????31??????????1

这里有几个看起来比较疑惑的地方：

①为什么这里要用min函数
②Keep这个东西是干什么的
③fist/last是干什么的
④dense_rank和dense_rank()有什么不同，能换成rank吗？

首先解答一下第一个问题：min函数的作用是用于当存在多个First/Last情况下保证返回唯一的记录。假如我们去掉会有什么样的后果呢？

SQL>?select?keep?(dense_rank?first?order?by?sum(customer_sales)?desc)?first,?
??2???????????? keep?(dense_rank?last?order?by?sum(customer_sales)?desc)?last
??3????from?user_order
??4???group?by?customer_id;
select?keep?(dense_rank?first?order?by?sum(customer_sales)?desc)?first,
????????????????????????*
ERROR?at?line?1:
ORA-00907:?missing?right?parenthesis

接下来看看第2个问题：keep是干什么用的？从上面的结果我们已经知道Oracle对排名的结果只“保留”2条数据，这就是keep的作用。告诉Oracle只保留符合keep条件的记录。

那么什么才是符合条件的记录呢？这就是第3个问题了。dense_rank是告诉Oracle排列的策略，first/last则告诉最终筛选的条件。

第4个问题：如果我们把dense_rank换成rank呢？

SQL>?select?min(region_id)
??2??????????keep(rank?first?order?by?sum(customer_sales)?desc)?first,
??3?????????min(region_id)
??4??????????keep(rank?last?order?by?sum(customer_sales)?desc)?last
??5????from?user_order
??6???group?by?region_id;
select?min(region_id)
*
ERROR?at?line?1:
ORA-02000:?missing?DENSE_RANK

四、按层次查询：

现在我们已经见识了如何通过Oracle的分析函数来获取Top/Bottom N，第一个，最后一个记录。有时我们会收到类似下面这样的需求：找出订单总额排名前1/5的客户。

很熟悉是不？我们马上会想到第二点中提到的方法，可是rank函数只为我们做好了排名，并不知道每个排名在总排名中的相对位置，这时候就引入了另外一个分析函数ntile，下面我们就以上面的需求为例来讲解一下：

SQL>?select?region_id,
??2?????????customer_id,
??3?????????ntile(5)?over(order?by?sum(customer_sales)?desc)?tile
??4????from?user_order
??5???group?by?region_id,?customer_id;

?REGION_ID?CUSTOMER_ID???????TILE
----------?-----------?----------
????????10??????????31??????????1
?????????9??????????25?????????? 1
????????10??????????26??????????1
?????????6???????????6??????????? 1?????????
?????????8??????????18?????????? 2
?????????5???????????2??????????? 2
?????????9??????????23?????????? 3
?????????6???????????9??????????? 3
?????????7??????????11?????????? 3
?????????5???????????3??????????? 4
?????????6???????????8??????????? 4
?????????8??????????16?????????? 4
?????????6???????????7??????????? 5
????????10??????????29??????????5
?????????5???????????1??????????? 5

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ntile函数为各个记录在记录集中的排名计算比例，我们看到所有的记录被分成5个等级，那么假如我们只需要前1/5的记录则只需要截取TILE的值为1的记录就可以了。假如我们需要排名前25%的记录(也就是1/4)那么我们只需要设置ntile(4)就可以了。

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转载自：http://www.blogjava.net/pengpenglin/archive/2008/06/27/211019.html

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