Question

我有以下查询，它返回一个项目的多个结果。我只想获得最新日期（最新）的1项。我目前正在获得每个有条目的日期的结果。

SELECT DISTINCT 
    [a].[Name]
  , [a].[IsGhostConvey]
  , [b].[AttorneyName]
  , [b].[IsGhostConveyAttorney]
  , [c].[InstitutionID]
  , [f].[Base_CreatedDate] AS 'DateLatestAccountOpened'
FROM 
    [Intermediary] [a]
INNER JOIN 
    [Attorney] [b] ON [a].[Name] = [b].[AttorneyName]
INNER JOIN 
    [IntermediaryInstitution] [c] ON [c].[IntermediaryID] = [a].[ID]
INNER JOIN  
    [IntermediaryBranch] [d] ON [d].[IntermediaryID] = [a].[Id]
INNER JOIN 
    [Matter] [e] ON [e].[BranchID] = [d].[Id]
INNER JOIN 
    [Account] [f] ON [f].[MatterID] = [e].[ID]
ORDER BY 
    [a].[Name], [f].[Base_CreatedDate] DESC;

我想我错过了WHERE，GROUP BY或HAVING声明。

我的结果如下：

Name    IsGhostConvey   AttorneyName    IsGhostConveyAttorney   InstitutionID   DateLatestAccountOpened
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2017-10-05 08:46:47.573
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2017-05-25 09:00:24.853
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2017-03-03 10:07:01.753
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2016-12-15 08:47:59.393
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2016-11-09 14:55:29.763
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2016-11-09 14:54:11.877
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2016-05-30 08:43:47.993
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   NULL
Adriaan Booyens Attorneys   1   Adriaan Booyens Attorneys   1   1   2018-02-05 12:25:17.233
Adriaan Booyens Attorneys   1   Adriaan Booyens Attorneys   1   1   2017-06-14 10:25:42.687

但我想这样：

Name    IsGhostConvey   AttorneyName    IsGhostConveyAttorney   InstitutionID   DateLatestAccountOpened
Adlam Attorneys - Benoni    1   Adlam Attorneys - Benoni    1   2   2017-10-05 08:46:47.573
Adriaan Booyens Attorneys   1   Adriaan Booyens Attorneys   1   1   2018-02-05 12:25:17.233

Answer 1

您可以使用row_number()来执行如下复杂查询：

with t as (
    SELECT a.Name, a.IsGhostConvey, b.AttorneyName, b.IsGhostConveyAttorney, c.InstitutionID, f.Base_CreatedDate AS 'DateLatestAccountOpened'
    FROM Intermediary a
    INNER JOIN Attorney b
    ON a.Name = b.AttorneyName
    INNER JOIN IntermediaryInstitution c
    ON c.IntermediaryID = a.ID
    INNER JOIN IntermediaryBranch d
    ON d.IntermediaryID = a.Id
    INNER JOIN Matter e
    ON e.BranchID = d.Id
    INNER JOIN Account f
    ON f.MatterID = e.ID
    )
select t.*
from (select t.*,
      row_number() over (partition by name order by DateLatestAccountOpened desc) as seqnum
      from t
     ) t
where seqnum = 1
order by Name, Base_CreatedDate DESC;

Answer 2

您可以使用

GROUP BY (name)

为每个名称获得1个结果

SQL Server筛选查询结果仅包含具有最新日期

2 个答案: