将多个 SQL 查询组合成单个结果答案

【问题标题】：Combine multiple SQL queries into single result将多个 SQL 查询组合成单个结果
【发布时间】：2021-11-25 04:50:11
【问题描述】：

我正在尝试将 2 个查询的结果放入一个结果集中。我正在使用 SQL Server 2019 Express。

这是我正在处理的数据：

表Sales

SaleDate	SaleAmt	CustomerID
11/1/2021	500	123
11/1/2021	100	234
11/1/2021	300	345
11/2/2021	500	456
11/2/2021	100	567
11/2/2021	200	678

表Customers

CustomerID	CustomerName
123	Jon Doe
234	Jane Doe
456	Bob Doe
678	Jim Doe

查询 #1：

select sales.saledate, sum(sales.saleamt) as 'Total Sales from All' 
from Sales 
group by sales.saledate

查询 #2：

select sales.saledate, sum(sales.saleamt) as 'Total Sales from Customers'
from Sales 
where sales.customerid in (select customerid from customers) 
group by sales.saledate

这是我想要的结果：

SaleDate	Total Sales from All	Total Sales from Customers
11/1/2021	900	600
11/2/2021	800	700

【问题讨论】：

标签： sql-server

【解决方案1】：

您可以在销售日期使用加入

select s1.saledate, All_Total AS 'Total Sales from All', CustomersTotal as 'Total Sales from Customers'
from (
         select sales.saledate, sum(sales.saleamt) as All_Total
         from Sales
         group by sales.saledate
     ) s1
         inner join
     (
         select sales.saledate, sum(sales.saleamt) as CustomersTotal
         from Sales
         where sales.customerid in (select customerid from customers)
         group by sales.saledate
     ) s2 on s1.saledate = s2.saledate

【讨论】：

【解决方案2】：

您可以使用case 表达式将其组合在一个查询中。

select s.saledate, 
       sum(s.saleamt) as [Total Sales from All],
       sum(case when exists 
                     (
                         select * 
                         from   customers c 
                         where  c.customerid = s.customerid 
                     ) 
                then s.salesamt 
                end) as [Total Sales from Customers]
from   Sales s
group by s.saledate

【讨论】：

【解决方案3】：

您可以将LEFT JOIN 与条件聚合一起使用

select
  s.saledate,
  sum(s.saleamt) as [Total Sales from All],
  sum(case when c.customerid is not null then s.saleamt end) as [Total Sales from Customers]
from Sales s
left join customers c on s.customerid = c.customerid
group by s.saledate;

【讨论】：