VB6 ms访问sql查询区分计数两个表

Question

I need to output per a specific employee this data: employee_Id, employee_name, presence.year, presence.month, count(day), sum(work hours), sum(holiday hours), [work hours] - [holiday hours]. I used this query on vb6 project but it seems not working good:

SELECT
     Employees.Employee_ID AS ID
     , Employees.Full_Name AS Name
     , Presence.Yr AS Year
     , Presence.Mnth AS Month
     , SUM(IIF(Presence.Dy, 1, 0)) Work_Days
     , SUM(IIF(Presence.Hr, Presence.Hr, 0)) AS Work_Hours
     , SUM(IIF(Presence.Holidays, Presence.Holidays, 0)) AS Holidays_Hours
     , SUM(IIF(Presence.Hr, Presence.Hr, 0)) - SUM(IIF(Presence.Holidays, Presence.Holidays, 0) AS Total_Work_Hours
FROM Employees
INNER JOIN Presence ON Presence.Employee_ID = Employees.Employee_ID
WHERE Employees.Company = 'aCompany'
     AND Employees.Employee_ID = 'anEmmployee'
GROUP BY  Employees.Employee_ID, Employees.Full_Name, Presence.Yr, Presence.Mnth

Like this in the picture: http://oi60.tinypic.com/122c6xz.jpg

Can anyone help with this?

Answer 1

Here is your query with some adjustments:

SELECT
     Employees.Employee_ID AS ID
     , Employees.Full_Name AS Name
     , Presence.Yr AS Year
     , Presence.Mnth AS Month
     , SUM(IIF(Presence.Dy, 1, 0)) Work_Days
     , SUM(IIF(Presence.Hr, Presence.Hr, 0)) AS Work_Hours
     , SUM(IIF(Presence.Holidays, Presence.Holidays, 0)) AS Holidays_Hours
     , SUM(IIF(Presence.Hr, Presence.Hr, 0)) - SUM(IIF(Presence.Holidays, Presence.Holidays, 0) AS Total_Work_Hours
     , SUM(Presence.Hr) / Employees.HoursPerDay + Employees.TotalAnnualHolidays - Employees.DoneHolidays AS AvailableHolidays
FROM Employees
INNER JOIN Presence ON Presence.Employee_ID = Employees.Employee_ID
WHERE Employees.Company = 'aCompany'
     AND Employees.Employee_ID = 'anEmmployee'
GROUP BY  Employees.Employee_ID, Employees.Full_Name, Employees.HoursPerDay, Employees.TotalAnnualHolidays, Employees.DoneHolidays, Presence.Yr, Presence.Mnth

Can you test this new query and give me info about the result and the difference with what you're looking for?

Hope this will help.