如何使用 PostgreSQL 中的排名函数将在线订单与之前的几个网站访问相关联

Question

我正在努力创建一个数据库视图，将在线订单与用户之前访问网站相关联。这是一个电子商务网站，因此一个用户可能会多次访问和订购。

我已经在 user_id 加入了访问 table 和订单 table，并将最近的小于会话时间与订单时间相关联。现在，我希望说每次访问直到订单 #1 为“1”，然后在访问之后直到订单 #2 为“2”。此外，如果该特定用户没有 order_id，我想 return“0”。请参阅下面链接的屏幕截图以供参考。

我已经尝试使用 dense_rank，但它只对存在 order_id 的行进行排名。我要把这些队伍发扬光大

SELECT v.id AS visit_id,
    v.user_id,
    v.started_at AS visit_date,
    dense_rank() OVER (PARTITION BY v.user_id ORDER BY v.started_at) AS visit_number,
    dense_rank() OVER (PARTITION BY v.user_id ORDER BY o.id) AS order_number,
    o.id AS order_id,
    o.created_at AS order_date
   FROM visits v
     FULL JOIN orders o ON v.user_id = o.user_id AND v.started_at < o.created_at AND o.created_at < (( SELECT min(visits.started_at) AS min
           FROM visits
          WHERE visits.user_id = v.user_id AND visits.started_at > v.started_at)) AND (v.started_at + '24:00:00'::interval) > o.created_at
  GROUP BY v.id, v.user_id, v.started_at, o.id, o.created_at
  ORDER BY v.started_at;

Answer 1

使用 lag 检查前一行是否为非空行，以便将其标记为新的组开始。设置标志后，您可以使用运行总和来定义组。

SELECT T.*,
       1+SUM(FLAG) OVER(PARTITION BY user_id ORDER BY visit_date) AS order_number
FROM (
SELECT v.id AS visit_id,
    v.user_id,
    v.started_at AS visit_date,
    dense_rank() OVER (PARTITION BY v.user_id ORDER BY v.started_at) AS visit_number,
    o.id AS order_id,
    o.created_at AS order_date,
    --conditioncheck with lag
    case when lag(o.id) over(partition by v.user_id order by v.started_at) is not null then 1 else 0 end as flag
   FROM visits v
     FULL JOIN orders o ON v.user_id = o.user_id AND v.started_at < o.created_at AND o.created_at < (( SELECT min(visits.started_at) AS min
           FROM visits
          WHERE visits.user_id = v.user_id AND visits.started_at > v.started_at)) AND (v.started_at + '24:00:00'::interval) > o.created_at
    ) T

Answer 2

GROUP BY 似乎没有必要，但我会留下它。你基本上需要一个累计金额。

我会将特定订单之前的所有访问分配给订单号：

SELECT v.id AS visit_id, v.user_id,
       v.started_at AS visit_date,
       dense_rank() OVER (PARTITION BY v.user_id ORDER BY v.started_at) AS visit_number,
       dense_rank() OVER (PARTITION BY v.user_id ORDER BY o.id) AS order_number,
       o.id AS order_id,
       o.created_at AS order_date,
       count(o.id) over (partition by v.user_id order by v.started_at) as order_number
FROM visits v FULL JOIN
     orders o
     ON v.user_id = o.user_id AND
        v.started_at < o.created_at AND
        o.created_at < (SELECT min(visits.started_at)
                        FROM visits v2 
                        WHERE v2.user_id = v.user_id AND 
                              v2.started_at > v.started_at) AND
        (v.started_at + '24:00:00'::interval) > o.created_at
GROUP BY v.id, v.user_id, v.started_at, o.id, o.created_at
ORDER BY v.started_at;

我想这就是你想要的逻辑：

SELECT v.id AS visit_id, v.user_id,
       v.started_at AS visit_date,
       dense_rank() OVER (PARTITION BY v.user_id ORDER BY v.started_at) AS visit_number,
       dense_rank() OVER (PARTITION BY v.user_id ORDER BY o.id) AS order_number,
       o.id AS order_id,
       o.created_at AS order_date,
       MIN(o.order_number) OVER (PARTITION BY v.user_id ORDER BY v.started_at DESC) as order_number
FROM visits v FULL JOIN
     (SELECT o.*,
             ROW_NUMBER() OVER (PARTITION BY o.user_id ORDER BY o.id) as order_number
      FROM orders o
     ) o
     ON v.user_id = o.user_id AND
        v.started_at < o.created_at AND
        o.created_at < (SELECT min(visits.started_at)
                        FROM visits v2 
                        WHERE v2.user_id = v.user_id AND 
                              v2.started_at > v.started_at) AND
        (v.started_at + '24:00:00'::interval) > o.created_at
GROUP BY v.id, v.user_id, v.started_at, o.id, o.created_at
ORDER BY v.started_at;

它可能会在您想要 0 的地方生成 NULL，但是。

如何使用 PostgreSQL 中的排名函数将在线订单与之前的几个网站访问相关联

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