Question

我正在学习如何在SQL Server中使用Group By，我正在尝试编写一个查询，让我可以在Tickets之后的数字表中从inner joining获取所有信息它与其他表格BY SELECTED DATE。

我的表格如下：

第一张表格让我更详细地了解每张门票中售出的物品：

 DATE               | Ticket_ID | Items_Sold | Item_Name      | Client_ID
2018-04-17 00:00:00      1           3          Watch              1
2018-04-17 00:00:00      1           2          Snapnack Hat       1       
2018-04-17 00:00:00      2           1          Arizona Tea        1    
2018-04-17 00:00:00      2           2          Wristband          1
2018-04-18 00:00:00      3           2          Pants              1
2018-04-18 00:00:00      3           2          Wristband          1
2018-04-18 00:00:00      4           1          Pants              1
2018-04-18 00:00:00      4           1          Wristband          1

第二张表格给出了每张门票内容和付费总额的基本描述。

 DATE_Sale                  | Ticket_ID | Employee_ID | Total | Client_ID
 2018-04-17 00:00:00              1          1          150       1
 2018-04-17 00:00:00              2          1          60        1             
 2018-04-18 00:00:00              3          2          200       1
 2018-04-18 00:00:00              4          2          100       1

所以，如果我想知道2018-04-17 00:00:00和2018-04-18 00:00:00之间发生了什么，请按Client_ID和DATE分组

我想要的结果如下：

DATE_Sale             | Tickets | Total_Items_Sold | Total_Paid | Client_ID
2018-04-17 00:00:00        2             8             210          1      
2018-04-18 00:00:00        2             4             300          1

问题在于，当我使用'内部联接'来获取销售的总项目时，所有总和都给了我错误的结果，因为它似乎是从重复的行添加，因为表1包含相同Ticket_ID和相同日期

我目前的查询如下：

SELECT B.DATE_Sale, 
       B.Client_ID, COUNT(B.Ticket_ID) as Tickets, 
       SUM(A.Items_Sold) as Total_Items_Sold, 
       SUM(B.Total_Paid) as Total_Paid
FROM Table_A as A
INNER JOIN Table_B as B
ON A.Client_ID = B.Client_ID
AND B.DATE_sales = A.DATE_Sales
Group by B.DATE_sales, A.Client_ID

Answer 1

好吧，您可以使用subquery与相关代替

select a.DATE as DATE_Sale, COUNT(DISTINCT a.Ticket_ID) as Tickets,
       sum(a.Items_Sold) as Total_Items_Sold,
       (select sum(Total) from Table_B 
        where DATE_Sale = a.DATE and 
              Client_ID = a.Client_ID) as Total_Paid, a.Client_ID
from Table_A as a
group by a.DATE, a.Client_ID;

另外，您可以使用subquery并执行join

select a.DATE as DATE_Sale, COUNT(DISTINCT a.Ticket_ID) as Tickets,
          sum(a.Items_Sold) as Total_Items_Sold,
          b.Total_Paid, a.Client_ID   
from Table_A as a
inner join (select DATE_Sale, Client_ID, 
                   sum(Total) as Total_Paid
             from Table_B 
             group by DATE_Sale, Client_ID) as b 
ON b.DATE_Sale = a.DATE and 
   b.Client_ID = a.Client_ID
group by a.DATE, a.Client_ID;

Answer 2

;with cte as
(select DATE_Sale,client_id, count(distinct ticket_id ) as tickets, sum(items_sold) as total_items_sold
 from table_a group by client_id, DATE_Sale),
 cte2 as
(select client_id,DATE_Sale, count(distinct ticket_id) as tickets, sum(total) as total_paid
         from table_b group by client_id, DATE_Sale) 
select a.DATE_Sale_is,a.tickets, a.total_items_sold, b.total_paid,a.client_id from cte a join cte2 b
on a.client_id = b.client_id
and a.DATE_Sale_is = b.DATE_Sale

Answer 3

我会用两级聚合来做这件事，因为它捕获了数据应该如何填充。第一级是票证级别，第二级是您想要的汇总级别：

select t1.date, count(*) as num_tickets, sum(num_items) as num_items,
       sum(t2.total_paid) as total_paid
from (select date, ticket_id, client_id, count(*) as num_items, 
      from table1
      group by date, ticket_id, client_id
     ) t1 join
     table2 t2
     on t1.ticket_id = t2.ticket_id and
        t2.date_sale = t1.date and t2.client_id = t1.client_id
group by t1.date;

COUNT(DISTINCT)通常是避免多级聚合的捷径。在这种情况下，您确实需要聚合并使用大量子查询/ CTE来解决问题只会使查询更加复杂。

使用SUM时的问题，因为它计算重复的ID行

3 个答案: