比较一周加入的两个表的计数 - SQL

Compare counts of two tables joined on week - SQL

我是 SQL 的新手。我有两张桌子。我想在第一个中计算每周一件事的出现次数,在第二个中每周计算另一件事的出现次数,然后比较它们。 我已经在两个单独的图表中有了用于计数的代码,但似乎无法加入它们。

我的第一个数:

select 
  date_part('week',Table2.date at time zone 'utc' at time zone 'Europe/Paris') as week, 
  count(Table2.issue_solved) as count2
from Table2
where date is not null
group by week
order by week asc

我的第二个计数

select 
  date_part('week',Table1.activity_date at time zone 'utc' at time zone 'Europe/Paris') as week, 
  count(distinct Table1.activity_id) as count1
from Table1
left join X
  on Y1 = Y2
left join W
  on A1 = A2
  and B1 = B2

where activity_dimensions.type in ('Training')
  and acquisition_opportunity_dimensions.product_family = 'EHR'
  and activity_dimensions.country = 'fr'
  and activity_date::date >= date_trunc('[aggregation]', [daterange_start])
  and activity_date::date <= [daterange_end]
  and activity_date::date <= current_date

group by week
order by count_training_meetings desc

我试图在一周内通过连接将第一个代码连接到第二个代码中,但我似乎无法完成这项工作。

有什么想法吗?

不确定 periscope 是否允许 full join,但是如果您的第一个数据集(查询)中有几个星期没有出现在第二个数据集中,反之亦然,您应该在为了取回所有东西。

coalesce 旨在获取它识别为非空的第一个值。

标准的sql,应该是这样的

select
  coalesce(q1.week, q2.week) as week,
  count1,
  count2
from 
  (
    select 
      date_part('week',Table2.date at time zone 'utc' at time zone 'Europe/Paris') as week, 
      count(Table2.issue_solved) as count2
    from Table2
    where date is not null
    group by week
  ) q1
  full join
  (
    select 
      date_part('week',Table1.activity_date at time zone 'utc' at time zone 'Europe/Paris') as week, 
      count(distinct Table1.activity_id) as count1
    from Table1
    left join X
      on Y1 = Y2
    left join W
      on A1 = A2
      and B1 = B2
    where activity_dimensions.type in ('Training')
      and acquisition_opportunity_dimensions.product_family = 'EHR'
      and activity_dimensions.country = 'fr'
      and activity_date::date >= date_trunc('[aggregation]', [daterange_start])
      and activity_date::date <= [daterange_end]
      and activity_date::date <= current_date
    group by week
  ) q2
    on q1.week = q2.week

正如我在之前的评论中告诉您的那样,如果您的数据中存在不同年份的周,那么混合这些周可能是错误的,但这只是一个建议