在列列表中使用子查询

Question

我想创建一个查询，以计算在过去的7,14和28天内创建了多少条记录。我的结果会返回类似于：

7Days 14Days 28Days 
21  35  56 

我知道如何为每个时间安排例如7天，但我可以在一个查询中捕获所有三个？

select count(*) from Mytable 
where Created > DATEADD(day,-8, getdate()) 

Answer 1

而且不漂亮，但不依赖于子（表/列名来自AdventureWorks中）。如果它属于你的标准内的情况下语句返回1，否则为0 - 那么你只总结的结果：

select sum(case when datediff(day, modifieddate, getdate()) <= 7 
       then 1 else 0 end) as '7days', 
     sum(case when datediff(day, modifieddate, getdate()) > 7 
        and datediff(day, modifieddate, getdate()) <= 14 
       then 1 else 0 end) as '14days', 
     sum(case when datediff(day, modifieddate, getdate()) > 14 
        and datediff(day, modifieddate, getdate()) <= 28 
       then 1 else 0 end) as '28days' 
from sales.salesorderdetail 

编辑：更新DATEDIFF函数 - 它是书面的方式，它会返回一个负数（假设修改日期过去）导致所有项目属于第一种情况。感谢舍甫琴科M代表指出了这一点

Answer 2

这是不是世界上最漂亮的代码，但它的伎俩。尝试从三个子查询中进行选择，每个范围一个。

select * from 
(select COUNT(*) as Cnt from Log_UrlRewrites where CreateDate >= DATEADD(day, -7, getdate())) as Seven 
inner join (select COUNT(*) as Cnt from Log_UrlRewrites where CreateDate >= DATEADD(day, -14, getdate())) as fourteen on 1 = 1 
inner join (select COUNT(*) as Cnt from Log_UrlRewrites where CreateDate >= DATEADD(day, -28, getdate())) as twentyeight on 1 = 1 

Answer 3

select 
(
    select count(*) 
    from Mytable 
    where Created > DATEADD(day,-8, getdate()) 
) as [7Days], 
(
    select count(*) 
    from Mytable 
    where Created > DATEADD(day,-15, getdate()) 
) as [14Days], 
(
    select count(*) 
    from Mytable 
    where Created > DATEADD(day,-29, getdate()) 
) as [28Days] 

Answer 4

SELECT 
    [7Days] = COUNT(CASE UtmostRange WHEN 7 THEN 1 END), 
    [14Days] = COUNT(CASE UtmostRange WHEN 14 THEN 1 END), 
    [28Days] = COUNT(CASE UtmostRange WHEN 28 THEN 1 END) 
FROM (
    SELECT 
    *, 
    UtmostRange = CASE 
     WHEN Created > DATEADD(day, -8, GETDATE()) THEN 7 
     WHEN Created > DATEADD(day, -15, GETDATE()) THEN 14 
     WHEN Created > DATEADD(day, -29, GETDATE()) THEN 28 
    END 
    FROM Mytable 
) s