我有这样的查询
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.product, tc.sub-product;
现在，在我的查询中，我想获得每个类别的前10个产品（按销售量排名最高），并且每个类别我需要前5个子类别（按销售量排名前列）
您可以假设问题陈述是这样的：
按销售获得每个类别的前10个产品，按销售获得每个5个子产品。
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，因为结果确实很大。我正在阅读有关诸如rank之类的oracle分析功能的信息。有人可以帮我使用解析函数修改此查询。任何其他方法也可以起作用。
我指的是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..我想在此停留两天:(
可能有一些原因不使用分析函数，而是 仅 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, rf, rfm, 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, rf, rfm, 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.rf, 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_rf2, rownum_rfm;
我使用rownum而不是等级，因此您永远不会获得联系，换句话说，联系将是随机决定的。如果数据不够密集（前10个产品中的任何5个子产品少于- 可能显示其他产品的子产品），这也将不起作用。但是，如果数据密集（建立的数据库很大），则查询应该可以正常工作。
下面的数据进行了两次传递，但在每种情况下均返回正确的结果。同样，这是一个无联系等级查询。
select am, rf, rfm, 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, ttc.rfm, tg.count_rf, tg.rownum_rf, count(*) count_rfm, 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, ttc.rf, 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, tg.value, 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.rf, ttc.rfm, tg.count_rf, tg.rownum_rf ) X -- filter where the subproduct rank is in top 5 where rownum_rfm <= 5 order by am, rownum_rf, 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)
