在 dense_rank window 函数中添加计数条件 SQL

Question

我有一个功能可以计算您访问了多少次以及您是否已转换。我想要的是 dense_rank 重新开始计数，如果有转换：

SELECT
          uid,
          channel,
          time,
          conversion,
          dense_rank() OVER (PARTITION BY uid ORDER BY time asc) as visit_order
FROM table

当前table输出：此客户 (uid) 在访问 18 时进行了一次转化，现在我希望 dense_rank 的 visit_order 计数从同一客户的 0 重新开始，直到它达到下一次非转化-无效的。

Answer 1

看到这个（我不喜欢“试试这个”）：

SELECT
   id,
   ts,
   conversion,
   -- SC,
   ROW_NUMBER() OVER (PARTITION BY id,SC) R
FROM (
   SELECT
      id,
      ts,
      conversion,
      -- COUNT(conversion) OVER (PARTITION BY id, conversion=0 ORDER BY ts ) CC,
      SUM(CASE WHEN conversion=1 THEN 1000 ELSE 1 END) OVER (PARTITION BY id ORDER BY ts ) - SUM(CASE WHEN conversion=1 THEN 1000 ELSE 1 END) OVER (PARTITION BY id ORDER BY ts )%1000 SC
   FROM sample
   ORDER BY ts
   ) x
ORDER BY ts;

DBFIDDLE

输出：

id	ts	conversion	R
1	2022-01-15 10:00:00	0	1
1	2022-01-16 10:00:00	0	2
1	2022-01-17 10:00:00	0	3
1	2022-01-18 10:00:00	1	1
1	2022-01-19 10:00:00	0	2
1	2022-01-20 10:00:00	0	3
1	2022-01-21 10:00:00	0	4
1	2022-01-22 10:00:00	0	5
1	2022-01-23 10:00:00	0	6
1	2022-01-24 10:00:00	0	7
1	2022-01-25 10:00:00	1	1
1	2022-01-26 10:00:00	0	2
1	2022-01-27 10:00:00	0	3

在 dense_rank window 函数中添加计数条件 SQL

Add a counting condition into dense_rank window Function SQL

sql

dense-rank