Question

我试图计算可通过不同渠道联系的客户。以下代码将仅提供可通过SMS联系的客户的统计数据。

with grouping as (
select  distinct                        
case when sms_correct_flag = 'Y' then 'Y' else 'N' end as smsable,                      
case when email_correct_flag = 'Y' then 'Y' else 'N' end as emailable,                  
case when address_correct_flag = 'Y' then 'Y' else 'N' end as dmable,           
contact_key                     
from raw.contacts                   
)

select count(distinct contact_key)
from grouping 
where smsable = 'Y';

我希望最终找到一张包含频道＆＃39;频道＆＃39;作为专栏，＆＃39;电子邮件＆＃39;，＆＃39;短信＆＃39;＆＃39; DM＆＃39;作为行，并填写他们各自的客户数量。

这可能是一个计数（情况......）但是当我们检查一个与我们的列不同的列的情况时，无法弄清楚如何完成这项工作正在数。

任何帮助表示赞赏！

Answer 1

由于您只有三个频道，因此您可以使用UNION ALL来生成所需的结果：

with grouping as (
    select
        MAX(case when sms_correct_flag = 'Y' then 1 else 0 end) as smsable
    ,   MAX(case when email_correct_flag = 'Y' then 1 else 0 end) as emailable
    ,   MAX(case when address_correct_flag = 'Y' then 1 else 0 end) as dmable
    ,   contact_key
    from raw.contacts
    group by contact_key
)
select 'sms' as channel, SUM(smsable) as cust_count from grouping
union all
select 'email' as channel, SUM(emailable) as cust_count from grouping
union all
select 'dm' as channel, SUM(dmable) as cust_count from grouping

注意：我不知道亚马逊的redshift是否内置了枢轴设施。如果确实如此，那么你可能有一个比这个穷人的支点实施更好的方法。

Answer 2

这是你在找什么？

select channel,                        
       sum(case when sms_correct_flag = 'Y' then 1 else 0 end) as smsable,                      
       sum(case when email_correct_flag = 'Y' then 1 else 0 end) as emailable,                  
       sum(case when address_correct_flag = 'Y' then 1 else 0 end) as dmable          
from raw.contacts  
group by channel;

SQL：如果列y = z，则计算列x

2 个答案: