1

我目前有一个生成 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(为了得到上面的输出):

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]  

我想做的是加入另一个名为 [Home_Days_table] 的表。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

我正在寻找上述查询以填充 [Home_Days_Table] 中的 [WFH 日期] 字段,用于在 [Results Review].[Review Date] 和 [Results Review].[Reviewed By] 之间匹配的每条记录与 [Home_Days_Table].[审阅者] 和 [Home_Days_Table].[WFH 日期]。我想显示上面原始 SQL 中的所有记录,并附加一个满足此条件的列。

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,但要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]   

上述查询回答了原始问题,但我想在 [WFH 日期] 顶部添加另一列,称为 [Summer Days]。这个想法与原始问题相同,但我正在寻找查询以填充 [Home_Days_Table] 中的 [WFH Date] 字段和 [Home_Days_Table] 中的 [Summer Days] 字段,用于匹配的每条记录在 [结果审核].[审核日期] 和 [结果审核].[审核人] 与 [Home_Days_Table].[审核人] 和 [Home_Days_Table].[WFH 日期] 和 [结果审核].[审核日期] & [Results Review].[Reviewed By] with the [Home_Days_Table].[Reviewed By] & [Home_Days_Table].[Summer Days]。

有人可以帮忙吗?

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1 回答 1

1

就像是 .. ?

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]  
于 2012-06-18T19:08:52.703 回答