在 Redshift 查询中为 max(date) 函数指定意外事件

Question

我试图确定客户在一个日历周内何时未登录，他们最后一次登录是什么时候。起始数据集如下所示：

User_Name   Login_Date  Week_Ending
Bobisaur    1/1/2019    1/5/2019
Bobisaur    1/3/2019    1/5/2019
Bobisaur    3/5/2019    3/9/2019
Bobisaur    3/24/2019   3/30/2019
Bobisaur    4/1/2019    4/6/2019

现在以1/12/2019 等结束的那一周，客户Bobisaur 将无法登录我想要做的是最终得到一个如下所示的数据集：

User_Name   Week_Ending (for weeks with no login)   Days Since Last Login (as of week ending date)
Bobisaur    1/12/2019    7 
Bobisaur    1/19/2019    14 
Bobisaur    1/26/2019    21 
Bobisaur    2/2/2019     28 
Bobisaur    2/9/2019     35 
Bobisaur    2/16/2019    42 
Bobisaur    2/23/2019    49 
Bobisaur    3/2/2019     56 
Bobisaur    3/16/2019    11 
Bobisaur    3/23/2019    18 
Bobisaur    4/13/2019    12 

这将过滤到他们没有登录的用户的周列表中，并有一个列显示自他们上次登录以来的天数，截至该特定的周末日期。

我能够获得没有登录活动的 Week_Ending 日期，但是，我一直在计算“自上次登录以来的天数（截至周末日期）”。我尝试做的是使用(Week_Ending - max(Login_Date))，然后使用max(Login_Date) <= Week_Ending 指定一个有子句。

但是，这基本上删除了 Week_Ending 值早于最大 Login_Date 的所有行。

任何帮助将不胜感激。谢谢！

Answer 1

生成的查询看起来很复杂，但还不错：

• intput_raw - 你在问题中粘贴的任何内容
• input - 将 login_data 与各自的 week_ending 交换
• cal - 带有连续星期六的日历表（您对 weekending 的定义）。在 Redshift 中，生成行的唯一方法是 SELECT。在这里，我通过将输入与自身交叉连接来生成 25 行，以获得 25 (5x5) 个连续的星期六。 CROSS JOIN 可以改为SELECT row_number() over () from arbitrary table limit 25。
• cal_usrs - 所有用户和所有星期六的餐桌
• output_raw - 计算的核心 - 首先是 JOIN cal_usrs 和 input，然后使用窗口函数获取 自上次登录后的天数。由于不可能对带有窗口函数结果的列进行过滤（并且您所需的输出没有带有0 的行），因此最终有一个SELECT。
• final SELECT - 只选择我们感兴趣的内容。

查询：

with input_raw as (
  select 'Bobisaur' as username, '1/1/2019'::date as login_date
  union all
  select 'Bobisaur', '1/3/2019'::date
  union all
  select 'Bobisaur', '3/5/2019'::date
  union all
  select 'Bobisaur', '3/24/2019'::date
  union all
  select 'Bobisaur', '4/1/2019'::date
), input as (
  select
         username,
         -- in your example weeks ends on saturday hence Monday + 5 days
         date_trunc('week', login_date) + interval '5 days' as week_ending
  from input_raw
), cal as (
  -- this will create a table with consecutive Saturdays
  select
         date_trunc('week', '12/1/2018'::date) + interval '5 days'+ 7 * row_number() over () as week_ending
  -- can be changed to 'from arbitrary table limit 25' or whatever time window you wish
  from input_raw a cross join input_raw b --this will produce 25 rows

), cal_usrs as (
  select * from cal cross join (select distinct username from input) as u
  -- this is very important - you want to have all weeks with all users
), output_raw as (
  select cal_usrs.username,
         cal_usrs.week_ending,
         max(input.week_ending)
             over (partition by cal_usrs.username order by cal_usrs.week_ending rows between unbounded preceding and current row ) as last_login_week,
         extract('days' from cal_usrs.week_ending - last_login_week) as days_since_last_login
  from input
         right join cal_usrs using (username, week_ending)
)
select
  username,
  to_char(week_ending, 'MM/DD/YYYY') as week_anding,
  days_since_last_login
from output_raw
where days_since_last_login <> 0 -- your example did not contain 0 rows
order by week_ending

结果看起来像（我认为您在 3 月 16 日之后的示例中计算错误）：

    username    week_ending days_since_last_login
    Bobisaur    01/12/2019  7
    Bobisaur    01/19/2019  14
    Bobisaur    01/26/2019  21
    Bobisaur    02/02/2019  28
    Bobisaur    02/09/2019  35
    Bobisaur    02/16/2019  42
    Bobisaur    02/23/2019  49
    Bobisaur    03/02/2019  56
    Bobisaur    03/16/2019  7
    Bobisaur    03/30/2019  7
    Bobisaur    04/13/2019  7
    Bobisaur    04/20/2019  14
    Bobisaur    04/27/2019  21
    Bobisaur    05/04/2019  28
    Bobisaur    05/11/2019  35
    Bobisaur    05/18/2019  42
    Bobisaur    05/25/2019  49