如何按不同条件对列求和，然后按日期分组答案

【问题标题】：how to SUM a column by different condition then group by date如何按不同条件对列求和，然后按日期分组
【发布时间】：2021-05-13 22:12:12
【问题描述】：

我尝试将列（小计）的值汇总到不同列（总计、拒绝、实际）中，这些列根据列的条件（状态）进行区分。

实际表（ode_orders）

Delivery Date	Subtotal	State
1/2/2021	150.00	Delivery Completed
2/2/2021	45.00	Rejected by business
2/2/2021	45.00	Delivery Completed
2/2/2021	87.00	Rejected by business
3/2/2021	45.00	Delivery Completed
3/2/2021	70.00	Delivery Completed

我希望它成为新表：

Delivery Date	Total Sales	Rejected	Actual
1/2/2021	150	0	150
2/2/2021	177	132	45
3/2/2021	155	0	155

我尝试使用此查询：

SELECT  delivery_date, 
  SUM(subtotal) AS TotalSale,
  SUM(subtotal) WHERE state NOT LIKE %Completed% AS TotalSale
FROM ode_orders
GROUP BY delivery_date;

结果

Error Code: 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE state NOT LIKE %Completed% AS TotalSale
FROM ode_orders
GROUP BY deliver' at line 3

【问题讨论】：

标签： mysql sql group-by sum

【解决方案1】：

使用条件聚合

 SELECT delivery_date, SUM(subtotal) AS TotalSale,
 SUM(case when State='Rejected by business' then subtotal else 0 end) as Rejected ,
SUM(case when State='Delivery Completed' then subtotal else 0 end) as actual 
 from table_name group by delivery_date

【讨论】：

谢谢你！另一个问题是如何按日期添加订单总数？就像我们已经得到了实际的总量，但是如果我们也计算一下总订单的数量会更好。例如，总共 10 个订单 RM150。
如果有助于标记为已接受，只需计数（*）以及已回答的查询即可
我已经得到了所有的总数。但是当我按“拒绝”条件计算时，它会计算所有行，我想排除“已完成”条件。 @zaynul

【解决方案2】：

select * 
,(totalSales-Rejected) Actual
from
  (
  select deliveryDate 
  ,sum(subTotal) totalSales 
  ,isnull((
  select sum(subTotal)  from ode_orders sa where sa.deliveryDate=a.deliveryDate and 
  [state]='Rejected by business'
  ),0) Rejected
  from ode_orders a
  group by deliveryDate
) mt

【讨论】：