排除總和大於1的記錄

Question

我有查詢返回的已訂閱頻道xyz或所有其他頻道的客戶的詳細信息。 為了生成此結果，我使用以下查詢：

select customerID
,sum(case when channel='xyz' then 1 else 0 end) as 'xyz Count'
,sum(case when channel<>'xyz' then bundle_qty else 0 end) as 'Other'
From temptable

所以我的問題是，我如何排除已訂閱2個頻道的客戶，其中一個是xyz，另一個是另一個頻道。

Answer 1

select customerID
      ,sum(case when channel='xyz' then 1 else 0 end) as 'xyz Count'
      ,sum(case when channel<>'xyz' then bundle_qty else 0 end) as 'Other'
From temptable
group by customerID
having sum(case when channel= 'xyz' then 1 else 0 end) > 0
   and sum(case when channel<>'xyz' then 1 else 0 end) > 0

Answer 2

首先，您的查詢不正確。 它需要一個group by 。 其次，您可以使用having ：

select customerID,
       sum(case when channel = 'xyz' then 1 else 0 end) as xyz_Count,
       sum(case when channel<>'xyz' then bundle_qty else 0 end) as Other
From temptable
group by customerID
having count(*) = 2 and
       sum(case when channel = 'xyz' then 1 else 0 end) = 1;

如果客戶可以多次訂閱同一頻道，而您仍然只希望“ xyz”和另一個頻道，則：

having count(distinct channel) = 2 and
       (min(channel) = 'xyz' or max(channel) = 'xyz')