sql - PostgreSQL 发现最后三个登录日期之间的差异

Question

我想为每个客户获取最后三个登录日期，并找到在上次登录 (login3) 和上次登录 (login1) 之前的时间超过 4 天的客户。

“活动”表包含：

• 用户身份
• login_date 为 DATETIME 格式，但时间始终为 00:00:00
• （以及其他一些与问题字段无关的）

我尝试了几个查询，但没有一个能正常工作。

Answer 1

这是一种可以在 PostgreSQL 8.3 及更高版本中使用数组的解决方案。

生成测试数据。改变第二个参数generate_series()以添加更多活动记录：

create table activity (id serial primary key, user_id integer, login_date timestamp);
insert into activity (user_id, login_date)
  select * from
  (
   select round(random()*10)::integer as user_id, ('2012-01-01'::date + (round(random()*300))* '1 day'::interval) as login_date
   from
   (select generate_series(1,1000)) foo
  ) fooger order by login_date;

select * from activity;

查询出想要的数据：

--show last three login dates per user:
select user_id, login[1] as login1, login[2] as login2, login[3] as login3
from
(
 select user_id, array_agg(login_date) as login from
 (select * from activity order by user_id,login_date desc) foo
 group by user_id
)  foo;

--shake out those who haven't been visiting frequently enough
select user_id, login[1] as login1, login[2] as login2, login[3] as login3, (login[1] - coalesce(login[3],login[2],login[1]))::interval as diff
from
(
 select user_id, array_agg(login_date) as login from
 (select * from activity order by user_id,login_date desc) foo
 group by user_id
)  foo
where login[1] - coalesce(login[3],login[2],login[1])  > '4 days'::interval;

Answer 2

我使用并简化了@Joshua 提供的设置：

CREATE TEMP TABLE activity (id serial primary key, user_id integer
                                                 , login_date timestamp);
INSERT INTO activity (user_id, login_date)
SELECT * FROM  (
   SELECT round(random()*10)::int AS user_id
        , ('2012-01-01 0:0'::timestamp + random() * interval '365 days') AS ts
   FROM   generate_series(1,1000)
   ) g
ORDER  BY ts;

您可以使用自 PostgreSQL 8.4 起可用的窗口函数：

SELECT user_id, login1, login3, (login1 - login3) AS time_span
FROM   (
   SELECT user_id, login_date
         ,first_value(login_date)      OVER w  AS login1
         ,COALESCE(lead(login_date, 2) OVER w 
                  ,lead(login_date)    OVER w) AS login3 
   FROM   activity
   WINDOW w AS (PARTITION BY user_id ORDER BY login_date DESC)
    ) x
WHERE  login_date = login1
AND    (login1 - login3) > interval '4d';

阅读 IMO 更容易，但在快速测试中，@Joshua 的查询速度快了约 30% 。

• 只有一个条目的用户永远不会有资格。
• 对于只有两个条目的用户，使用倒数第二个而不是倒数第三个。

除此之外，如果您的时间戳的时间部分始终是00:00:00您可能需要考虑使用date列而不是timestamp.

Answer 3

为了完整性：NAIVE 版本（查询计划显示 3 个 CTE 的三个单独的子计划；这很糟糕）（递归 CTE 也应该是可能的 ;-)

WITH l3 AS (
        SELECT a3.id, a3.user_id, a3.login_date
        FROM   activity a3
        WHERE NOT EXISTS ( SELECT *
                FROM activity nx
                WHERE nx.user_id = a3.user_id
                AND nx.login_date > a3.login_date
                )
        )
, l2 AS (
        SELECT a2.id, a2.user_id, a2.login_date
        FROM   activity a2
        JOIN l3 ON l3.user_id = a2.user_id AND l3.login_date > a2.login_date
        WHERE NOT EXISTS ( SELECT *
                FROM activity nx
                WHERE nx.user_id = a2.user_id
                AND nx.login_date > a2.login_date
                AND nx.login_date < l3.login_date
                )
        )
, l1 AS (
        SELECT a1.id, a1.user_id, a1.login_date
        FROM   activity a1
        JOIN l2 ON l2.user_id = a1.user_id AND l2.login_date > a1.login_date
        WHERE NOT EXISTS ( SELECT *
                FROM activity nx
                WHERE nx.user_id = a1.user_id
                AND nx.login_date > a1.login_date
                AND nx.login_date < l2.login_date
                )
        )
SELECT l1.user_id
        ,l1.id AS ii1, l1.login_date AS d1
        ,l2.id AS ii2, l2.login_date AS d2
        ,l3.id AS ii2, l3.login_date AS d3
FROM l1
JOIN l2 ON l2.user_id = l1.user_id
JOIN l3 ON l3.user_id = l1.user_id
WHERE l3.login_date - l1.login_date > '4 days'::INTERVAL
        ;