Question

我已经尝试但我无法弄清楚这一点。我有一个表事务（transaction_ID，transaction_Person_ID，Transaction_Date等）。我想要的是返回去年每周超过3笔交易的所有transaction_person_ID。这意味着我必须检查1-1-10到7-10-10，看看有人在那一周有超过3笔交易，然后是2-1-10到8-10-10等等。到目前为止我写的是这个

WITH Dates AS (
        SELECT
         [Date] = CONVERT(DATETIME,'01/01/2010')
        UNION ALL SELECT
         [Date] = DATEADD(DAY, 1, [Date])
        FROM
         Dates
        WHERE
         Date < '12/31/2010'
)

SELECT transaction_person_Id FROM transactions
JOIN DATES
ON transactions.transaction_date = dates.date
where transactions.Transaction_Date between dateadd(DAYOFYEAR,-7,dates.date) and dates.date
group by transaction_person_Id
having count(transaction_person_ID) >= 4
OPTION (MAXRECURSION 2000)

由于

Answer 1

我目前还没有ms-sql的实例，所以我不能测试它，但是一旦不可避免的语法错误得到纠正，它应该做你想做的事情

select 
    transaction_person_ID
from
(
    select 
        transaction_person_Id, 
        count(transaction_person_id), 
        datepart(wk,Transaction_date)
    from
        transactions
    Where
        Transaction_date > dateadd(d,-datepart(dy,getdate())+1,getdate())
    group by 
        transaction_person_Id, 
        datepart(wk,Transaction_date)
    having
        count(transaction_person_id) >3
) as foo
group by
   transaction_person_id
having
   count(transaction_person_id) >= datepart(wk,getdate)

希望有所帮助！

Answer 2

看一下这个链接

http://www.sqlservercentral.com/articles/T-SQL/71571/

返回每组的前X行，作者：Dave Ballantyne

T-SQL选择返回结果一段时间

2 个答案: