Question

我需要您按日期订购查询，然后是否可以简化： - ）

此查询将返回当天每周的条目数。

以下是查询：

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -7), 103) as periode_join, COUNT(u.usr_ID) as cptu 
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -7) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), 0)

UNION

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -14), 103) as periode_join, COUNT(u.usr_ID) as cptu 
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -14) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -7)   

UNION

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -21), 103) as periode_join, COUNT(u.usr_ID) as cptu 
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -21) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -14)

UNION

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -28), 103) as periode_join, COUNT(u.usr_ID) as cptu 
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -28) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -21)   

 UNION

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -35), 103) as periode_join, COUNT(u.usr_ID) as cptu  
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -35) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -28)   

UNION

SELECT CONVERT(VarChar(50), DATEADD(day, DATEDIFF(day, 0, GETDATE()), -42), 103) as periode_join, COUNT(u.usr_ID) as cptu 
FROM [USR_USER] u 
INNER JOIN [USI_USER_SITE] s ON u.USR_ID = s.USR_ID
WHERE u.[USR_JOINED_DT] >=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -42) and u.[USR_JOINED_DT] <=  DATEADD(day, DATEDIFF(day, 0, GETDATE()), -35)
ORDER BY periode_join desc

结果如下：

periode_join    cptu
28/05/2013  8740
25/06/2013  9773
18/06/2013  8212
11/06/2013  6644
04/06/2013  9420
02/07/2013  7868

非常感谢:-D

Answer 1

如果你有SQL Server 2005及更高版本，你可以尝试这样的事情：

;with CTE_Data as
(
  select
        u.usr_id,
        cast(dateadd(dd, -(datediff(dd, u.usr_joined_dt, getdate()) / 7 + 1) * 7, getdate()) as date) as periode_join
    from usr_user as u 
        inner join usi_user_site as s on u.usr_id = s.usr_id
)
select periode_join, count(usr_id) as cptu
from CTE_Data
group by periode_join
order by periode_join asc

SQL FIDDLE EXAMPLE

要按周拆分数据，我们必须在getdate（）和表格中的日期 - datediff(dd, u.usr_joined_dt, getdate())之间取得差异，然后我们必须获得整周的数量 - / 7，然后我们必须减去整数周+ 1。

因此，例如，如果日期= '2013/06/30'，则天数= 9，整周数= 1，我们必须减去1 + 1周，因此我们得到'2013/06/25'等等

按周获得结果

1 个答案: