如何將單個日期分組爲幾個月

Question

我試圖按月來總結這些數據的結果...而不是個別日期。我該怎麼做？

SELECT DISTINCT 
    CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101)) AS Appended_SignInDate, 
    GO.GeographicLocationDescription, 
    COUNT(DISTINCT acd.EmployeeID) as Total_FTL 

FROM  
    dbACD.Detail.vwRockwellEventAgentPerformance AS ACD 

LEFT OUTER JOIN  
    dbEmployee.Summary.vwEmployeeHistory AS EH 
    ON acd.employeeID = eh.EmployeeID 
    AND ACD.SignOutDateTime Between eh.StartDate and eh.EndDate 

LEFT OUTER JOIN  
    dbEmployee.Config.vwGeographicLocation AS GO 
    ON EH.GeographicLocationID = GO.GeographicLocationID 


WHERE 
    ACD.SignOutDateTime BETWEEN '2012-06-01' AND '2013-03-31' 
-- AND ACD.SignoutReasonCode ='4' 
    AND GO.GeographicLocationDescription is not null 

GROUP BY  

    CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101)), 
    GO.GeographicLocationDescription 


ORDER BY  
    CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101)), 
    GO.GeographicLocationDescription 

Answer 1

DatePart（）函數應該爲此工作。

http://msdn.microsoft.com/en-us/library/ms174420.aspx

SELECT DISTINCT 
    DATEPART(month, CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101))) AS Appended_SignInDate 
    , GO.GeographicLocationDescription, COUNT(DISTINCT acd.EmployeeID) as Total_FTL 
FROM dbACD.Detail.vwRockwellEventAgentPerformance AS ACD 
LEFT OUTER JOIN dbEmployee.Summary.vwEmployeeHistory AS EH 
    ON acd.employeeID = eh.EmployeeID 
    AND ACD.SignOutDateTime Between eh.StartDate and eh.EndDate 
LEFT OUTER JOIN dbEmployee.Config.vwGeographicLocation AS GO 
    ON EH.GeographicLocationID = GO.GeographicLocationID 
WHERE ACD.SignOutDateTime BETWEEN '2012-06-01' AND '2013-03-31' 
-- AND ACD.SignoutReasonCode ='4' AND GO.GeographicLocationDescription is not null 
GROUP BY 
    DATEPART(month, CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101))) 
    , GO.GeographicLocationDescription 

ORDER BY 
CONVERT (DATETIME, CONVERT (VARCHAR (10), SignInDateTime, 101)), GO.GeographicLocationDescription