摘要：
1.连续7天登陆但未下单的用户 select c.user_id, count(c.cum) as cum_01 from (select a.user_id,a.time01 - row_number() over(partition by a.user_id order_by a.time01 阅读全文
posted @ 2021-08-01 18:17 隐灰子 阅读(71) 评论(0) 推荐(0)
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摘要：
1.连续7天登陆但未下单的用户 select c.user_id, count(c.cum) as cum_01 from (select a.user_id,a.time01 - row_number() over(partition by a.user_id order_by a.time01 阅读全文
posted @ 2021-08-01 18:17 隐灰子 阅读(71) 评论(0) 推荐(0) |
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