计算具有...的行数 SQL

Question

我必须计算 UserId 是否被列出一次或多次，如果一次将 +1 设置为 "new users"，否则将“+1”设置为 "returning users"。

我做过吗：

select 
Count(distinct [UserId]) as 'Unique users'
from [TelemetryData] 
where [DiscountId] = '8CAEA860-6766-43E2-9280-27AFE7FDF82E' and [EventName] = 'DiscountClick'

/* returning */
select
count(Id) as 'Returning users'
from [TelemetryData] 
where [DiscountId] = '8CAEA860-6766-43E2-9280-27AFE7FDF82E' and [EventName] = 'DiscountClick'
group by [UserId]
having count(Id) > 1

/* returning */
select
count(*) as 'New users'
from [TelemetryData] 
where [DiscountId] = '8CAEA860-6766-43E2-9280-27AFE7FDF82E' and [EventName] = 'DiscountClick'
group by [UserId]
having count(*) = 1

我需要像第一个查询一样计算 "returning" 和 "new" 用户查询中的总行数。怎么做？

Answer 1

您可以使用两个级别的聚合来执行此操作。计算 UserId 级别的计数，然后使用该信息获得所需的计数：

select sum(case when cnt > 1 then 1 else 0 end) as ReturningUsers,
       sum(case when cnt = 1 then 1 else 0 end) as NewUsers
from (select UserId, count(*) as cnt
      from [TelemetryData]
      where [DiscountId] = '8CAEA860-6766-43E2-9280-27AFE7FDF82E' and [EventName] = 'DiscountClick'
      group by UserId
     ) u

计算具有...的行数 SQL

Count number of rows with having... SQL

sql

sql-server

having