SQL 查询以常量格式返回聚合，即使类别丢失

Question

SQL 服务器是否可以以常量格式返回表？

假设我们有以下原始数据：

DATE       |  CATEGORY  |  VALUE
---------------------------------
01.01.2022 | Category 1 |   10
01.01.2022 | Category 1 |   20
01.01.2022 | Category 1 |   33
01.01.2022 | Category 3 |   15
03.01.2022 | Category 1 |   10
03.01.2022 | Category 2 |   20
03.01.2022 | Category 3 |   50
              (...)

所需的 output 将是：

DATE       |  CATEGORY  |  VALUE
---------------------------------
01.01.2022 | Category 1 |    63
01.01.2022 | Category 2 |     0
01.01.2022 | Category 3 |    15
02.01.2022 | Category 1 |     0
02.01.2022 | Category 2 |     0
02.01.2022 | Category 3 |     0
03.01.2022 | Category 1 |    10
03.01.2022 | Category 2 |    20
03.01.2022 | Category 3 |    50
              (...)

请注意，在期望的结果中，原始数据中缺少存在的日期，并且当原始数据中的给定日期不存在类别时，VALUE 的总和为 0。

Answer 1

-- Contiguous dates table
DECLARE @dates TABLE(dt date) ;   
DECLARE @dateFrom date;
DECLARE @dateTo date;

select @dateFrom = (Select DateAdd(day, -1, Min(date)) from Agg);
select @dateTo = (Select Max(date) from Agg);

-- Query:


WHILE(@dateFrom < @dateTo)
BEGIN
   SELECT @dateFrom = DATEADD(day, 1,@dateFrom)
   INSERT INTO @dates 
   SELECT @dateFrom
END

-- Category table 
DECLARE @categories TABLE(category nvarchar(20)) ;
insert into @categories values ('Category 1'),('Category 2'),('Category 3');

-- This cte helps in creating the constant output required 
with cte1 as (
select dt, category from 
@dates cross join @categories

) 
select cte1.dt as [Date], cte1.category, Sum(coalesce(yourTableName.value,0)) as Value
from cte1 left join yourTableName
on cte1.dt = yourTableName.[Date] and cte1.category = yourTableName.category
group by cte1.dt, cte1.category
order by cte1.dt, cte1.category

Answer 2

WITH  ctedate AS
(
    SELECT d= v2.d * 10 + v1.d
    FROM (VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9)) v1(d)
    CROSS JOIN (VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9)) v2(d)
)

Select b.date, a.category, sum(coalesce(a.value,0)) from 
yourtablename a 
cross join
 (SELECT DATEADD(DAY, ctedate.d, '2022-01-01') date
FROM ctedate
ORDER BY ctedate.d) b 
group by b.date, a.category

Answer 3

这是一个使用cte解决方案的脚本：

1. 计算最小日期和最大日期

2. 然后使用递归制作一个日历（cte1），其中包含（最小日期）和（最大日期）之间的所有日期

3. 获取所有类别的列表 => cte2

4. 在 cte1 和 cte2 之间进行交叉连接==> cte3

5. 在 cte3 和表数据 (@mytable) 之间进行左连接，将 Null 值替换为 0

   将@mytable 声明为表（日期日期，类别 varchar(50)，值 int）

   插入@mytable 值 ('01/01/2022','Category 1',10), ('01/01/2022','Category 1',20), ('01/01/2022','Category 1',33), ('01/01/2022','类别 3',15), ('01/03/2022','类别 1',10), ('01/03/2022','类别 2',20), ('01/03/2022','类别 3',50);

   将@mindate 声明为日期，@maxdate 声明为日期

   select @mindate=min(date),@maxdate=max(date) from @mytable;

   与 cte1 为 (select @mindate mydate union all select dateadd(day,1,mydate) from cte1 where dateadd(day,1,mydate) <= @maxdate), cte2 as (select distinct(category) from @mytable), cte3 as (select mydate,category from cte1 cross join cte2), cte4 as (select mydate date,cte3.category,isnull(value,0) value from cte3 left outer join @mytable t on cte3.mydate=t.date and cte3.类别=t.category ) select * 来自 cte4