汇总案例

Question

是否可以为case语句提供2个聚合函数 我需要为客户的不同标志获取数量和销售量

基本查询：

Select count(distinct invoices), sum(tot_sal_amt) from stores

尝试：

    Select date,case count(invoices)
when (sum(case when custom_flag='guest' then tot_sal_amt  end)then count(invoices) end 
when (sum(case when custom_flag='non-guest' then tot_sal_amt  end)then count(invoices) end 
when (sum(case when custom_flag is null then tot_sal_amt  end)then count(invoices) end
from stores

任何帮助表示赞赏。

谢谢

Answer 1

如果可以使用以下格式，则可以：

select date,
       custom_flag,
       (case 
        when sum(case when custom_flag='guest' then tot_sal_amt end) then count(distinct invoices) 
        when sum(case when custom_flag='non-guest' then tot_sal_amt end) then count(distinct invoices) 
        when sum(case when custom_flag is null then tot_sal_amt  end) then count(distinct invoices) 
       end) as cnt_invoices,
       (case 
        when custom_flag='guest' then sum(tot_sal_amt)
        when custom_flag='non-guest' then sum(tot_sal_amt) 
        when custom_flag is null then sum(tot_sal_amt)
       end) as sum_tot_sal_amt      
  from stores
 group by date, custom_flag;

问题发生于

1. case count(invoices) when其中 count(invoices)应该被删除。
2. sum语句前面的多余的左括号。

但是您无需为您的情况使用case..when语句，因为在所有情况下结果都是相等的，因此使用以下命令就足够了：< / p>

 select date,
       custom_flag,
       count(distinct invoices) as cnt_invoices,
       sum(tot_sal_amt) as sum_tot_sal_amt 
  from stores
 group by date, custom_flag;  

Rextester Demo

Answer 2

在同一case语句中可以有两个或多个聚合函数，但不能返回多个值。

未测试伪代码。

Select  date
       ,Count(distinct invoices)
       ,Count(Case when custom_flag='guest'     then invoices
                   when custom_flag='non-guest' then invoices
                   when custom_flag is null then invoices end) as cnt_invoices
       ,Sum(Case when custom_flag='guest' then tot_sal_amt
                 when custom_flag='non-guest' then tot_sal_amt
                 when custom_flag is null then tot_sal_amt end) as sum_sal_amt
from stores
group by date

Answer 3

我很确定您想要条件聚合：

Select date, 
       count(distinct invoiceid) as num_invoices,
       count(distinct case when custom_flag = 'guest' then invoiceid end) as guest_invoices,
       count(distinct case when custom_flag = 'non-guest' then invoiceid end) as nonguest_invoices,
       count(distinct case when custom_flag is null then invoiceid end) as null_invoices,
       sum(total_sal_amt) as total_sales,
       sum(case when custom_flag = 'guest' then total_sal_amt end) as guest_total_sal_amt,
       sum(case when custom_flag = 'non-guest' then total_sal_amt end) as nonguest_total_sal_amt,
       sum(case when custom_flag is null then total_sal_amt end) as null_total_sal_amt
from stores
group by date;

或更简单地说：

Select date, custom_flag,
       count(distinct invoiceid) as num_invoices,
       sum(total_sal_amt) as total_sales,
from stores
group by date, custom_flag;