Sql Access-查询以提取每个唯一opportunity的最新条目

从名为

tblNotes

的表中，我需要为每个

opportunityid

获取最新的

notedata

以下是

tblNotes

中的数据

以下是我希望数据的样子：

我对SQL了解不多，但我发现我需要一个内部查询。我所做的阅读表明Access将

notesdate

存储为日期和时间代码

这是我拼凑的代码

SELECT t1.*
FROM tblNotes AS t1 INNER JOIN 
    (SELECT opportunityid, notesdata, MAX(notesdate) AS MaxDate 
     FROM tblnotes GROUP BY opportunityid, notesdata)  
    AS t2 ON (t1.opportunityid=t2.opportunityid) AND  (t1.notesdate=t2.MaxDate);

下面是我得到的：

正如你所看到的…甚至不接近。我想基本结构就藏在那里的某个地方……也许吧

谢谢，

我想你想要：

select n.*
from tblNotes as n
where n.notesId = (select top 1 n2.notesId
                   from tblNotes as n2
                   where n2.opportunityId = n.opportunityId
                   order by n2.notesDate desc, n2.notesId desc
                  );

假设

notesid

是主键，则每个

opportunityId

返回一行，即使最近的日期有多个注释。如果希望所有注释都在最近的日期，请使用：

select n.*
from tblNotes as n
where n.notesDate = (select max(t2.notesDate)
                     from tblNotes as n2
                     where n2.opportunityId = n.opportunityId
                    );

我想你想要：

select n.*
from tblNotes as n
where n.notesId = (select top 1 n2.notesId
                   from tblNotes as n2
                   where n2.opportunityId = n.opportunityId
                   order by n2.notesDate desc, n2.notesId desc
                  );

假设

notesid

是主键，则每个

opportunityId

返回一行，即使最近的日期有多个注释。如果希望所有注释都在最近的日期，请使用：

select n.*
from tblNotes as n
where n.notesDate = (select max(t2.notesDate)
                     from tblNotes as n2
                     where n2.opportunityId = n.opportunityId
                    );

---

我认为这是干净的：从输入创建一个带有oppty id和每个oppty id的max notes date的内联搜索表，并在较小表的两列上将该表内部连接回输入：

DROP TABLE input;                                                                  
CREATE TABLE 
input(notesid,opportunityid,notesdate,notesuser,notesdata) AS 
          SELECT  5,0,DATE '2020-07-14',3,'kkk'
UNION ALL SELECT  3,1,DATE '2020-07-14',2,'another test to run'
UNION ALL SELECT  1,1,DATE '2020-07-14',2,'This is just a test. Next action is..'
UNION ALL SELECT  9,1,DATE '2020-07-27',4,'test'
UNION ALL SELECT 12,1,DATE '2020-07-28',1,'another day in paradise'
UNION ALL SELECT 13,1,DATE '2020-08-14',1,'another test .. maybe this time'
UNION ALL SELECT  2,6,DATE '2020-07-14',3,'another test to see how it works'
;

SELECT
  i.*
FROM input i
INNER JOIN (
  SELECT
    opportunityid
  , MAX(notesdate) AS notesdate
  FROM input
  GROUP BY opportunityid
) m
  ON m.opportunityid=i.opportunityid
 AND m.notesdate    =i.notesdate
-- out  notesid | opportunityid | notesdate  | notesuser |            notesdata            
-- out ---------+---------------+------------+-----------+---------------------------------
-- out        2 |             6 | 2020-07-14 |         3 | another test to see how it works
-- out        5 |             0 | 2020-07-14 |         3 | kkk
-- out       13 |             1 | 2020-08-14 |         1 | another test .. maybe this time

---

我认为这是干净的：从输入创建一个带有oppty id和每个oppty id的max notes date的内联搜索表，并在较小表的两列上将该表内部连接回输入：

DROP TABLE input;                                                                  
CREATE TABLE 
input(notesid,opportunityid,notesdate,notesuser,notesdata) AS 
          SELECT  5,0,DATE '2020-07-14',3,'kkk'
UNION ALL SELECT  3,1,DATE '2020-07-14',2,'another test to run'
UNION ALL SELECT  1,1,DATE '2020-07-14',2,'This is just a test. Next action is..'
UNION ALL SELECT  9,1,DATE '2020-07-27',4,'test'
UNION ALL SELECT 12,1,DATE '2020-07-28',1,'another day in paradise'
UNION ALL SELECT 13,1,DATE '2020-08-14',1,'another test .. maybe this time'
UNION ALL SELECT  2,6,DATE '2020-07-14',3,'another test to see how it works'
;

SELECT
  i.*
FROM input i
INNER JOIN (
  SELECT
    opportunityid
  , MAX(notesdate) AS notesdate
  FROM input
  GROUP BY opportunityid
) m
  ON m.opportunityid=i.opportunityid
 AND m.notesdate    =i.notesdate
-- out  notesid | opportunityid | notesdate  | notesuser |            notesdata            
-- out ---------+---------------+------------+-----------+---------------------------------
-- out        2 |             6 | 2020-07-14 |         3 | another test to see how it works
-- out        5 |             0 | 2020-07-14 |         3 | kkk
-- out       13 |             1 | 2020-08-14 |         1 | another test .. maybe this time

---

NotesID是一个PK。进一步检查后，第一块代码几乎满足了我的需要。删除1周围的括号后，查询将根据

opportunityid

返回结果，其中包含相关的

notesid

。我需要为没有相应“notesid”的任何

opportunityid

返回一个空条目。有什么想法吗？如果更合适的话，我会提出一个新问题。谢谢。回复到这是正确的解决方案，因为它确实回答了我提出的问题。NotesID是一个PK。进一步检查后，第一块代码几乎满足了我的需要。删除1周围的括号后，查询将根据

opportunityid

返回结果，其中包含相关的

notesid

。我需要为没有相应“notesid”的任何

opportunityid

返回一个空条目。有什么想法吗？如果更合适的话，我会提出一个新问题。谢谢。回到这是正确的解决方案，因为它确实回答了我提出的问题。