Question

我试图根据过去18个月内的访问次数将客户与“延期”商家联系起来，其中决胜局是最近的访问日期。我在决胜局时遇到了一些麻烦。如果有两个记录根据某个MemberID的访问次数排名为1，我想在记录上将IsFirst位列设置为1，并为该MemberID设置MAX（EncounterDate）。我应该怎么做呢？

Answer 1

这可能对您有所帮助......这是基于现有emp表的Oracle查询。我认为在发布问题时创建结构是个好主意。替换第一个选择更新等...：更新您的表设置您的日期= max_date（在我的示例中为max_hire_date）WHERE your_field IN（选择我的示例中的最大日期）AND rnk = 1和rno = 1

SELECT * FROM 
 (  
 SELECT deptno
      , ename
      , sal
      , RANK() OVER (PARTITION BY deptno ORDER BY sal desc) rnk 
      , ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY sal desc) rno 
      , MAX(hiredate) OVER (PARTITION BY deptno ORDER BY deptno) max_hire_date
   FROM emp_test
  WHERE deptno = 20
 ORDER BY deptno
 )
 WHERE rnk = 1
   --AND rno = 1 -- or 2 or any other number...
/

SQL>

DEPTNO  ENAME   SAL    RNK  RNO HIREDATE    MAX_HIRE_DATE
-----------------------------------------------------------
 20     SCOTT   3000    1   1   1/28/2013   1/28/2013
 20     FORD    3000    1   2   12/3/1981   1/28/2013

Answer 2

以下SQL获取您想要的信息，假设某些表的结构：

select c.*, NumVisits, MaxVisitDate, MaxFirstVisitDate
       (select count(*)
        from visits v
        where v.customerid = c.customerid and 
              v.visi
from customers c join
     (select customerid,
             sum(case when visitdate between getdate() - 365*1.5 and getdate()
                      then 1 else 0
                  end) as NumVisits,
             max(visitdate) as MaxVisitDate,
             max(case when IsFirst = 1 then visitdate end) as MaxFirstVisitDate
      from visits
      group by customerid
     ) v
     on c.customerid = v.customerid

根据这些信息，您可以将逻辑放在一起，以实现您的目标。当MaxVisitDate = MaxFirstVisitDate时，该位在最近的日期设置。

您的更新问题的答案是这样的：

update visits
    set IsFirst = 1
    where exists (select max(encounterDate)
                  from visits v2
                  where v2.customer_id = visits.customer_id
                  having max(encounterDate) = visits.encounterDate)
                 )

基于日期的SQL排名

2 个答案: