MS Access SQL - >计算表1中的记录,执行表2中的条件,但显示所有记录

时间:2012-06-18 18:53:01

标签: sql ms-access join

我目前有一个产生3列的SQL语句:

    [Review Date] [Total Reviews] [Reviewed By]  

     10/24/2001         16            Jane
     10/24/2001         1             Bob
     10/24/2001         2             Chloe
     09/20/2001         17            Jane
     09/20/2001         34            Bob
     09/20/2001         86            Chloe
     02/04/2001         14            Jane
     02/04/2001         3             Bob
     02/04/2001         41            Chloe

SQL看起来像这样(以获得上述输出):

SELECT 

[Review Results].[Review Date], 
count([Review Results].[Reviewed By]) as [Total Reviews], 
[Review Results].[Reviewed By]

FROM 
[Review Results]

GROUP BY  [Review Results].[Review Date], [Review Results].[Reviewed By]


ORDER BY [Review Results].[Review Date]  

我想要做的是在[Home_Days_table]中加入另一个表。该表如下所示:

[Reviewed By]     [WFH Date]

   Jane           10/12/2011
   Jane           07/11/2010
   Bob            04/09/2002
   Jane           01/01/2007

我正在寻找以上查询来填充[Home_Days_Table]中的[WFH Date]字段,以便在[Results Review]之间匹配每条记录。[Review Date]& [结果回顾]。[评论者]与[Home_Days_Table]。[评论者]& [Home_Days_Table]。[WFH日期]。我想显示上面原始SQL中的所有记录,并附加满足此条件的列。

有人可以帮忙吗?

除了上述内容之外,我还希望按照给定的方式执行查询,但添加另一个日期字段。我使用的查询是:

SELECT 
   [Review Results].[Review Date], 
   count([Review Results].[Reviewed By]) as [Total Reviews], 
   [Review Results].[Reviewed By], 
   [Home_Days_table].[WFH Date]

FROM [Review Results]
LEFT JOIN [Home_Days_table]
ON  [Review Results].[Reviewed By]=[Home_Days_table].[Reviewed By] 
AND [Review Results].[Review Date]=[Home_Days_table].[WFH Date]

GROUP BY  
   [Review Results].[Review Date], 
   [Review Results].[Reviewed By],
   [Home_Days_table].[WFH Date]


ORDER BY [Review Results].[Review Date]   

以上查询回答了原始问题,但我想在[WFH日期]上添加另一个名为[Summer Days]的列。这个想法与原始问题相同,但我正在寻找查询现在填充[Home_Days_Table]中的[WFH Date]字段和[Home_Days_Table]中的[Summer Days]字段,用于每个匹配的记录在[结果审查]之间。[审查日期]& [结果回顾]。[评论者]与[Home_Days_Table]。[评论者]& [Home_Days_Table]。[WFH日期]和[结果审核]。[审核日期]& [结果回顾]。[评论者]与[Home_Days_Table]。[评论者]& [Home_Days_Table]。[夏日]。

有人可以帮忙吗?

1 个答案:

答案 0 :(得分:1)

像...一样的东西?

SELECT 
   [Review Results].[Review Date], 
   count([Review Results].[Reviewed By]) as [Total Reviews], 
   [Review Results].[Reviewed By], 
   [Home_Days_table].[WFH Date]

FROM [Review Results]
LEFT JOIN [Home_Days_table]
ON  [Review Results].[Reviewed By]=[Home_Days_table].[Reviewed By} 
AND [Review Results].[Review Date]=[Home_Days_table].[WFH Date]

GROUP BY  
   [Review Results].[Review Date], 
   [Review Results].[Reviewed By],
   [Home_Days_table].[WFH Date]


ORDER BY [Review Results].[Review Date]