Question

我有一张桌子，里面有商品，供应商和日期（还有一些其他东西）。我想找到所有独特的商品，供应商，日期在整个日期集中没有日期。从我的值中得出我在表中找到所有日期的集合，并且我想构建一个唯一商品，供应商和日期的列表，它们当前在集合中没有日期属性。例如：

我有数据：

COMMODITY|SUPPLIER|DATE
-----------------------
1        |1       |15.06.16
1        |2       |22.06.16
2        |1       |29.06.16

我希望查询返回：

COMMODITY|SUPPLIER|DATE
-----------------------
1        |1       |22.06.16
1        |1       |29.06.16
1        |2       |15.06.16
1        |2       |29.06.16
2        |1       |15.06.16
2        |1       |22.06.16

我该怎么做？

Answer 1

您可以使用cross join构建所有可能的组合来完成此操作，然后选择存在的组合：

select c.commodity, s.supplier, d.date
from (select distinct commodity from t) c cross join
     (select distinct supplier from t) s cross join
     (select distinct date from t) d left join
     t
     on t.commodity = c.commodity and
        t.supplier = s.supplier and
        t.date = s.date
where t.commodity is null;

Answer 2

使用Partition Outer Join这很容易做到：

WITH sample_data AS (SELECT 1 commodity, 1 supplier, to_date('15/06/2016', 'dd/mm/yyyy') dt FROM dual UNION ALL
                     SELECT 1 commodity, 2 supplier, to_date('22/06/2016', 'dd/mm/yyyy') dt FROM dual UNION ALL
                     SELECT 2 commodity, 1 supplier, to_date('29/06/2016', 'dd/mm/yyyy') dt FROM dual),
             dts AS (SELECT DISTINCT dt FROM sample_data)
SELECT sd.commodity,
       sd.supplier,
       dts.dt
FROM   dts
       LEFT OUTER JOIN sample_data sd PARTITION BY (sd.commodity, sd.supplier) ON (sd.dt = dts.dt)
WHERE  sd.dt IS NULL;

 COMMODITY   SUPPLIER DT
---------- ---------- -----------
         1          1 22/06/2016
         1          1 29/06/2016
         1          2 15/06/2016
         1          2 29/06/2016
         2          1 15/06/2016
         2          1 22/06/2016

这也意味着您只需要引用该表两次 - 一次获取唯一日期列表，然后再次将表加入您感兴趣的日期列表中，因此它应该相对较高

Answer 3

with t as(
 select 1 as COMMODITY, 1 as SUPPLIER, '15.06.16' as "date" from DUAL
 union all
 select 1,2, '22.06.16' from DUAL
 union all
 select 2,1, '29.06.16' from DUAL
)

select *
  from (select distinct commodity, supplier from t),
       (select distinct "date" from t)
MINUS
select * from t

构造当前集合中不存在的值列表

3 个答案: