sql - 如何使用 postgresql 计算排列？

Question

我有一个包含城市之间连接的大型数据库。每个连接都有一个开始和目的地城镇、一个开始日期以及该连接的价格。

我想为任何连接以及返回连接在 1-20 天之间的日期计算传出+返回连接的任何组合。然后为每个日期组合选择最优惠的价格。

例子:

表:

city_start,     city_end,   date_start,     price
Hamburg         Berlin      01.01.2016      100.00
Berlin          Hamburg     10.01.2016      112.00
Berlin          Hamburg     10.01.2016      70.00
Berlin          Hamburg     12.01.2016      50.00
Berlin          Hamburg     30.02.2016      20.00
Paris           Madrid      ...
Madrid          Paris
London          Paris

期望的结果:

Hamburg-Berlin-Hamburg, 01.01.2016, 10.01.2016, 170.00 (100+70)
Hamburg-Berlin-Hamburg, 01.01.2016, 12.01.2016, 150.00 (100+50)
...
(not Berlin-Hamburg on 30.02.2016 because it's >20 days from departure drive)
(not London-Paris, as there is no return Paris-London)

我可以通过以下方式获得可能的组合:

SELECT DISTINCT city_start, city_end, city_end, city_start from table

但是我现在如何计算它们的排列？

Answer 1

获取所有对的查询使用 join :

select tto.city_start, tto.city_end, tto.date_start, tfrom.date_end,
       (tto.price + tfrom.price) as price
from t tto join
     t tfrom
     on tto.city_end = tfrom.city_start and
        tto.city_start = tfrom.city_end and
        tfrom.date_start >= tto.date_start + interval '1 day' and
        tfrom.date_end <= tto.date_start + interval '20 day';

要获得最便宜的价格，请使用窗口函数:

select tt.*
from (select tto.city_start, tto.city_end, tto.date_start, tfrom.date_end,
             (tto.price + tfrom.price) as price,
             row_number() over (partition by tto.city_start, tto.city_end order by (tto.price + tfrom.price) asc) as seqnum
      from t tto join
           t tfrom
           on tto.city_end = tfrom.city_start and
              tto.city_start = tfrom.city_end and
              tfrom.date_start >= tto.date_start + interval '1 day' and
              tfrom.date_end <= tto.date_start + interval '20 day'
      ) tt
where seqnum = 1;