SQL:如果另一列不匹配,如何排除最大值

Question

我正在使用Oracle 10g.我在
SQL上的 研究- 如何选择具有最大值列的行
和
http://jan.kneschke.de/projects/mysql/groupwise-max/

解决如何处理max(行)然后找到其他max(行)但是,我还没有.

标准: ID最大值(enroll_date)的代码必须为"WHEAT".
我不希望代码为"WHEAT"的最长注册日期.只有当代码为最大注册日期的"WHEAT"时,我才想要该行.

问题:如何编写查询以满足条件？

这不起作用,因为它返回ID = 30,Code = WHEAT,其中Date 12/25/2001不是该ID的最大值(日期).

    select ID, code, max(enroll_date)
    from enrollment
    where  CODE = 'WHEAT'
    group by ID, code

这不起作用,因为它也返回ID = 30,Code = WHEAT,其中date = 12/25/2001.

    select ID, code, max(enroll_date)
    from enrollment
    group by ID, code
    having code='WHEAT'

这是我的桌子

ENROLLMENT
===========================
ID  CODE     ENROLL_DATE
--------------------------
01  WHEAT   <NULL>
01  WHEAT   12/21/2007
01  WHEAT   7/30/2009
30  WHEAT   12/25/2001
30  CHAFF   6/14/2010
72  WHEAT   8/20/2002
72  WHEAT   12/7/2007

 DESIRED RESULT 
 ID    CODE ENROLL_DATE
 ================================
  01     WHEAT  7/30/2009
  72     WHEAT  12/7/2007

Answer 1

这听起来像你想要的东西

SELECT id, 
       code,
       enroll_date
  FROM (SELECT id, 
               code,
               enroll_date,
               rank() over (partition by id 
                                order by enroll_date desc nulls last) rnk
          FROM enrollment)
 WHERE rnk = 1
   AND code = 'WHEAT'

解析函数RANK分配的最新ENROLL_DATE特定ID值1,下一个最近的ENROLL_DATE为ID2的值,等等,那么你可以参考排名,并在外部查询的代码.如果可能存在重复项,您可能更愿意使用DENSE_RANK或ROW_NUMBER解析函数

您也可以编写查询

SELECT id, 
       code,
       enroll_date
  FROM (SELECT id, 
               code,
               enroll_date,
               max(enroll_date) over (partition by id) max_enroll_date
          FROM enrollment)
 WHERE enroll_date = max_enroll_date
   AND code        = 'WHEAT'