组由分日期时间

Question

结果我试图让第一（最低）开始一个员工的日期：

SELECT a.StartDate, a.EmpID, b.fullname FROM Employees a 
LEFT JOIN dbo.info b ON a.EmpID = b.EmpID 
WHERE 
    type = 800 
    GROUP By EmpId 
    ORDER BY fullname 

结果与此类似：

StartDate | EmpId 
1992-12-01 00:00:00.000 | 7 
2014-01-01 00:00:00.000 | 7 
2015-01-01 00:00:00.000 | 7 
1992-12-01 00:00:00.000 | 10 
2014-01-01 00:00:00.000 | 10 
2015-01-01 00:00:00.000 | 10 
1992-01-01 00:00:00.000 | 16 
2014-01-01 00:00:00.000 | 16 
2015-01-01 00:00:00.000 | 16 

要得到的最低日每个员工我都试过这个，但是却导致了一个错误。

SELECT MIN(a.StartDate), a.EmpID, b.fullname FROM Employees a 
LEFT JOIN dbo.info b ON a.EmpID = b.EmpID 
WHERE 
    type = 800 
    GROUP By EmpId 
    ORDER BY fullname 

什么是正确的查询得到如下结果：

StartDate | EmpId 
1992-12-01 00:00:00.000 | 7 
1992-12-01 00:00:00.000 | 10 
1992-01-01 00:00:00.000 | 16 

Answer 1

SELECT a.StartDate, a.EmpID, b.fullname 
FROM Employees a 
LEFT JOIN dbo.info b ON a.EmpID = b.EmpID 
WHERE type = 800 
AND NOT EXISTS (SELECT 'a' 
       FROM Employees a2 
       WHERE a.EmpID = a2.EmpID 
       AND a2.StartDate < a.StartDate 
       ) 

Answer 2

需要每个EmpID一个结果行，所以这是你应该列组由：

SELECT MIN(StartDate), EmpID 
FROM  Employees 
WHERE type = 800 AND EmpID BETWEEN 1 AND 500 
GROUP BY EmpID 

Answer 3

SELECT MIN(StartDate), EmpID FROM Absences 
WHERE type = 800 
AND EmpID BETWEEN 1 AND 500 
GROUP BY StartDate 

你的团队是在错误的领域。

GROUP BY EmpID 

Answer 4

使用此查询， 这将停止重复数据

SELECT Distinct StartDate, EmpID FROM Employees 

Answer 5

SELECT DISTINCT a.EmpID, b.fullname, a.StartDate FROM Employees a 
LEFT JOIN dbo.info b 
ON a.EmpID = b.EmpID 
WHERE type = 800  
ORDER BY a.StartDate, b.fullname ASC