我有一个类似这样的查询
SELECT t.category,tc.product,tc.sub-product,count(*) as sales FROM tg t,ttc tc WHERE t.value = tc.value GROUP BY t.category,tc.sub-product;
现在在我的查询中,我希望获得每个类别的前十大产品(按销售额排名),对于每个类别,我需要前5个子类别(按销售额排名)
您可以将问题陈述假设为：
根据销售情况获得每个类别的前10个产品,每个产品按销售额排名前5个子产品.
>此处类别可以是书籍
>产品可以是Harry Porter的书
>子产品可以是HarryPorter系列5
样本输入数据格式
category |product |subproduct |Sales [count (*)] abc test1 test11 120 abc test1 test11 100 abc test1 test11 10 abc test1 test11 10 abc test1 test11 10 abc test1 test11 10 abc test1 test12 10 abc test1 test13 8 abc test1 test14 6 abc test1 test15 5 abc test2 test21 80 abc test2 test22 60 abc test3 test31 50 abc test3 test32 40 abc test4 test41 30 abc test4 test42 20 abc test5 test51 10 abc test5 test52 5 abc test6 test61 5 | | | bcd test2 test22 10 xyz test3 test31 5 xyz test3 test32 3 xyz test4 test41 2
输出将是“
top 5 rf for (abc) -> abc,test1(289) abc,test2 (140),abc test3 (90),abc test4(50),abc test5 (15) top 5 rfm for (abc,test1) -> test11(260),test12(10),test13(8),test14(6),test15(5) and so on
我的查询失败了,因为结果真的很大.我正在阅读像甲骨文这样的oracle分析函数.有人可以帮助我使用分析函数修改此查询.任何其他方法也可以工作.
我指的是这个http://www.orafaq.com/node/55.但是无法为此获得正确的SQL查询.
任何帮助将不胜感激.我喜欢在这上面坚持2天:(
解决方法
可能有理由不使用分析函数,而是单独使用分析函数：
select am,rf,rfm,rownum_rf2,rownum_rfm
from
(
-- the 3nd level takes the subproduct ranks,and for each equally ranked
-- subproduct,it produces the product ranking
select am,rownum_rfm,row_number() over (partition by rownum_rfm order by rownum_rf) rownum_rf2
from
(
-- the 2nd level ranks (without ties) the products within
-- categories,and subproducts within products simultaneosly
select am,row_number() over (partition by am order by count_rf desc) rownum_rf,row_number() over (partition by am,rf order by count_rfm desc) rownum_rfm
from
(
-- inner most query counts the records by subproduct
-- using regular group-by. at the same time,it uses
-- the analytical sum() over to get the counts by product
select tg.am,ttc.rf,ttc.rfm,count(*) count_rfm,sum(count(*)) over (partition by tg.am,ttc.rf) count_rf
from tg inner join ttc on tg.value = ttc.value
group by tg.am,ttc.rfm
) X
) Y
-- at level 3,we drop all but the top 5 subproducts per product
where rownum_rfm <= 5 -- top 5 subproducts
) Z
-- the filter on the final query retains only the top 10 products
where rownum_rf2 <= 10 -- top 10 products
order by am,rownum_rfm;
我使用的是rownum而不是rank,所以你永远不会得到关系,换句话说,关系会随机决定.如果数据不够密集(在前10个产品中的任何一个产品中少于5个副产品 – 它可能显示来自其他一些产品的副产品),这也不起作用.但是如果数据密集(大型建立的数据库),查询应该可以正常工作.
下面两次传递数据,但在每种情况下返回正确的结果.同样,这是一个无排名查询.
select am,count_rf,count_rfm,rownum_rf,rownum_rfm
from
(
-- next join the top 10 products to the data again to get
-- the subproduct counts
select tg.am,tg.rf,tg.count_rf,tg.rownum_rf,ROW_NUMBER() over (partition by tg.am,tg.rf order by 1 desc) rownum_rfm
from (
-- first rank all the products
select tg.am,tg.value,count(*) count_rf,ROW_NUMBER() over (order by 1 desc) rownum_rf
from tg
inner join ttc on tg.value = ttc.value
group by tg.am,ttc.rf
order by count_rf desc
) tg
inner join ttc on tg.value = ttc.value and tg.rf = ttc.rf
-- filter the inner query for the top 10 products only
where rownum_rf <= 10
group by tg.am,tg.rownum_rf
) X
-- filter where the subproduct rank is in top 5
where rownum_rfm <= 5
order by am,rownum_rfm;
列：
count_rf : count of sales by product count_rfm : count of sales by subproduct rownum_rf : product rank within category (rownumber - without ties) rownum_rfm : subproduct rank within product (without ties)
