Question

我有一个大型数据库，城市之间有连接。每个连接都有一个起始和目的地城镇，开始日期和该连接的价格。

我想计算任何连接的传出+返回连接的任何组合，以及返回连接在1-20天之间的日期。然后为每个日期组合选择最优价格。

示例：

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

但我现在如何计算他们的排列？

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，请使用窗口功能：

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;

Answer 2

这是没有row_number分区部分的解决方案：

SELECT
    a.city_start, a.city_end, b.city_end, a.date_start, b.date_start,
    min(a.price + b.price)
FROM
    flight AS a
    JOIN
    flight AS b ON a.city_start = b.city_end AND a.city_end = b.city_start
WHERE b.date_start BETWEEN a.date_start + 1 AND a.date_start + 20
GROUP BY a.city_start, a.city_end, b.city_end, a.date_start, b.date_start;

Answer 3

如果要包含其他列，请尝试以下操作：

SELECT
    a.city_start, a.city_end, b.city_end, a.date_start, b.date_start,
    a.price + b.price, a.car_name, b.car_name
FROM
    flight AS a
    JOIN
    flight AS b ON a.city_start = b.city_end AND a.city_end = b.city_start
    LEFT JOIN
    flight AS c ON
         a.city_start = c.city_start
         AND
         a.city_end = c.city_end
         AND
         a.date_start = c.date_start
         AND (
             a.price > c.price
             OR (
                 a.price = c.price
                 AND
                 a.id > c.id))
    LEFT JOIN
    flight AS d ON
         b.city_start = d.city_start
         AND
         b.city_end = d.city_end
         AND
         b.date_start = d.date_start
         AND (
             b.price > d.price
             OR (
                 b.price = d.price
                 AND
                 b.id > d.id))
WHERE
    b.date_start BETWEEN a.date_start + 1 AND a.date_start + 20
    AND
    c.id IS NULL
    AND
    d.id IS NULL;

如何使用postgresql计算排列？

3 个答案: