http://www.91linux.com/html/article/database/oracle/20081216/14775.html

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非常詳細的講述了分析函數的應用.

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部分例子代碼:

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1.統計分析數據.

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select * from (
?select o.cust_nbr customer,
???????? o.region_id region,
???????? sum(o.tot_sales) cust_sales,
???????? sum(sum(o.tot_sales)) over(partition by o.region_id) region_sales
??? from orders_tmp o
?? where o.year = 2001
?? group by o.region_id, o.cust_nbr
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?? ) where cust_sales/region_sales>0.2
?? order by region_sales desc

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2.排序. Rank，Dense_rank，Row_number

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select region_id, customer_id, sum(customer_sales) total,
????? rank() over(order by sum(customer_sales) desc) rank,
????? dense_rank() over(order by sum(customer_sales) desc) dense_rank,
?????? row_number() over(order by sum(customer_sales) desc) row_number
???? from user_order
?? group by region_id, customer_id;

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3.top/bottom n、first/last、ntile

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1.帶空值的排列

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?select region_id, customer_id,
?????????? sum(customer_sales) cust_total,
?????????? sum(sum(customer_sales)) over(partition by region_id) reg_total,
?????????? rank() over(partition by region_id
?????????????????????? order by sum(customer_sales) desc NULLS LAST) rank
????????? from user_order
???????? group by region_id, customer_id;

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綠色高亮處，NULLS LAST/FIRST告訴Oracle讓空值排名最后后第一。

注意是NULLS，不是NULL。

(經測試,在一般的排序語句中,也可以使用 NULLS LAST/FIRST 指定NULL值的排序位置.)

2.Top/Bottom N查詢

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【1】找出所有訂單總額排名前3的大客戶：
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;

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3.First/Last排名查詢

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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;

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4.按層次查詢

現在我們已經見識了如何通過Oracle的分析函數來獲取Top/Bottom N，第一個，最后一個記錄。有時我們會收到類似下面這樣的需求：找出訂單總額排名前1/5的客戶。

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

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