Question

希望查看是否可以比较特定用户（EDI）或其他任何人一天输入的订单数量。

我可以每天返回结果（但只能返回存在值的日子），但无法找到一种将所有三个结果组合在一起的方法（总计-通过EDI-其他所有人）。

非常感谢任何帮助。

select Date, count(Order_ID) 
from orders 
WHERE Date >=dateadd(day,datediff(day,0,GetDate())- 7,0) and [user] = 'EDI' 
      and customer = '9686'
GROUP BY Date, [user];
select Date, count(Order_ID) 
from orders 
WHERE Date >=dateadd(day,datediff(day,0,GetDate())- 7,0) and [user] <> 'EDI' 
      and customer = '9686'
GROUP BY Date, [user];
select Date, count(Order_ID) 
from orders 
WHERE Date >=dateadd(day,datediff(day,0,GetDate())- 7,0) 
      and customer = '9686'
GROUP BY Date, [user];

Answer 1

使用条件聚合：

select Date, sum(case when [user] = 'EDI' then 1 else 0 end) as cnt_edi,
       sum(case when [user] <> 'EDI' then 1 else 0 end) as cnt_non_edi,
       count(*) as total
from orders 
where Date >= dateadd(day, datediff(day, 0, GetDate()) - 7, 0) and customer = '9686'
group by Date;

每天的记录计数，按用户比较

1 个答案: