这似乎应该是一个经典的发票项目问题已经解决了,但也许我没有在我的搜索中使用正确的单词。
我正在运行这样的查询(这只是一个简单的例子,我的真实查询要复杂得多,但它会返回相同的结果):
select invoice.inv_num, item.name, item.qty
from invoice invoice, item
where invoice.inv_num = item.inv_num
order by invoice.inv_num
我需要生成一个项目编号列,该列对每个项目递增,但每个新发票编号从1开始。因此,例如,我需要最终结果看起来像这样:
inv_num item_num name qty
------- -------- ------------- ---
111 1 red widgets 10
111 2 blue widgets 5
222 1 green_widgets 7
222 2 red_widgets 16
222 3 black_widgets 10
333 1 blue_widgets 8
333 2 red_widgets 12
我们仍在使用Oracle 9i以防万一。
答案 0 :(得分:4)
您可以使用oracle rank或row_number分析函数(取决于您希望如何处理重复项/ euqally排名项)。
以下是将第4列item_number添加到查询中的方法:
select invoice.inv_num, item.name, item.qty ,
row_number() over (partition by inv_num order by qty desc) item_num
from invoice invoice, item
where invoice.inv_num = item.inv_num
order by invoice.inv_num
http://docs.oracle.com/cd/B19306_01/server.102/b14200/functions001.htm#i81407
答案 1 :(得分:3)
select invoice.inv_num,
item.name,
item.qty,
row_number() OVER(PARTITION BY invoice.inv_num order by item.qty desc) as item_num
from invoice invoice, item
where invoice.inv_num = item.inv_num
order by invoice.inv_num
row_number()
生成以1开头的数字..然后我们使用PARTITION BY
子句重新启动每个* INV_NUM *的序列。并用数量排序编号。
答案 2 :(得分:0)
假设发票的项目名称是唯一的,您可以使用rank()聚合函数:
create table TESTTABLE(inv_num NUMBER,iname VARCHAR2(100 CHAR),qty NUMBER);
insert into TESTTABLE values(111,'red widgets',10);
insert into TESTTABLE values(111,'blue widgets',5);
insert into TESTTABLE values(222,'green_widgets',7);
insert into TESTTABLE values(222,'red_widgets',16);
insert into TESTTABLE values(222,'black_widgets',10);
insert into TESTTABLE values(333,'blue_widgets',8);
insert into TESTTABLE values(333,'red_widgets',12);
commit;
select inv_num, iname, qty ,
rank() over (PARTITION BY inv_num ORDER BY iname) as item_num
from TESTTABLE
order by inv_num;