Question

我需要找到连续三个月发布三次或更多次的用户。我写了这个问题：

select count(id), owneruserid, extract(month from creationdate) as postmonth from posts
group by owneruserid, postmonth
having count(id) >=3
order by owneruserid, postmonth

我明白了：

count   owneruserid postmonth
36     -1            1
23     -1            2
45     -1            3
41     -1            4
18     -1            5
24     -1            6
31     -1            7
78     -1            8
83     -1            9
17     -1            10
88     -1            11
127    -1            12
3      6             11
3      7             12
4      8             1
8      8             12
4      12            4
3      12            5
3      22            2
4      22            4

（截短的）

哪个好。如何查询连续三次或更多，三个月或更长时间发布的用户？感谢。

Answer 1

这被称为群岛和差距问题，特别是它是具有日期范围的岛屿问题。你应该，

解决此问题。
将其标记为发送至dba.stackexchange.com

要解决这个问题，

如果前面的行与前一个mont不对应，则创建一个窗口为1的伪列
使用COUNT()
检查以确保该组的count(*)大于或等于三。

查询，

SELECT l.id, creationdaterange, count(*)
FROM (

  SELECT t.id,
    t.creationdate,
    count(range_reset) OVER (PARTITION BY t.id ORDER BY creationdate) AS creationdaterange
  FROM (
    SELECT id,
    creationdate,
    CASE
      WHEN date_trunc('month',creationdate::date)::date - interval '1 month' = date_trunc('month',lag(creationdate))::date OVER (PARTITION BY id ORDER BY creationdate)
      THEN 1
    END AS range_reset
    FROM post
    ORDER BY id, creationdate
  ) AS t;

) AS l
GROUP BY t.id, creationdaterange
HAVING count(*) >= 3;

如何在Postgresql中评估数据？

1 个答案: