计算工作天数

Question

你们是如此善于打理代码。我可以使用更好的SQL来告诉我每个员工每月工作的天数。每个员工每天可以进出几次，并且可以在午夜之前工作。如果他们在午夜工作，则计为工作2天。如果他们在午夜时分工作并在当天晚些时候进入，并在下一个午夜之前离开，那么这段时间自同一天以来就已经被计算在内。

这有效，但有没有更简单的方法？

IF OBJECT_ID ('dbo.ZTable1', 'U') IS NOT NULL 
    DROP TABLE dbo.ZTable1; 
GO 
CREATE TABLE dbo.ZTable1 ([EmployeeId] Numeric (5,0), [TimeIn] datetime, 
[TimeOut] datetime) 

INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 1,'2017-09- 
13 12:19','2017-09-14 00:01' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 1,'2017-09- 
14 12:15','2017-09-15 00:01' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 1,'2017-09- 
15 12:35','2017-09-16 00:01' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 1,'2017-09- 
16 07:56','2017-09-16 10:31' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 1,'2017-09- 
16 11:56','2017-09-16 16:31' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 2,'2017-09- 
13 15:26','2017-09-14 00:00' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 2,'2017-09- 
14 15:29','2017-09-15 00:00' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 2,'2017-09- 
15 15:27','2017-09-16 00:01' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 3,'2017-09- 
13 15:25','2017-09-14 00:01' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 3,'2017-09- 
14 15:25','2017-09-15 00:00' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 3,'2017-09- 
15 15:26','2017-09-16 00:00' 
INSERT INTO dbo.ZTable1 ([EmployeeId],[TimeIn],[TimeOut]) SELECT 3,'2017-09- 
16 06:55','2017-09-16 15:27' 

GO; 

With Step1 as ( --< Build temp table of in punch days 
Select [EmployeeId], DATEPART (DAY ,[TimeIn]) as WorkDay 
from dbo.ZTable1 
), 
Step2 as ( --< Build temp table of out punch days 
Select [EmployeeId], DATEPART (DAY ,[TimeOut]) as WorkDay 
from dbo.ZTable1 
), 
Step3 as ( --< merges both in and put punch tables 
Select 
Case when s1.[EmployeeId] is NULL then s2.[EmployeeId] else s1.[EmployeeId] end as Employee, 
case when s1.WorkDay is NULL then s2.WorkDay else s1.WorkDay end as WorkDate 
from Step1 s1 
full outer join Step2 s2 on s1.[EmployeeId] = s2.[EmployeeId] and s1.WorkDay = s2.WorkDay 
), 
Step4 as ( --< Organizes temp table 
Select Distinct Employee, WorkDate 
from Step3 
group by Employee, WorkDate 
) 
Select Employee, Count (Employee) as NumDays 
from Step4 
Where Employee > 0 
Group by Employee 
Order by Employee 


DROP TABLE dbo.ZTable1 

Output (Result) 
Employee NumDays 
1   4 
2   4 
3   4 

Answer 1

试试这个，没有热膨胀系数需要：

SELECT EmployeeID, COUNT(1) as NumDays 
FROM (
    SELECT EmployeeID, CONVERT(DATE,TimeIn) DistinctDays 
    FROM ZTable1 

    UNION 

    SELECT EmployeeID, CONVERT(DATE,TimeOut) 
    FROM ZTable1 
    ) A 
GROUP BY EmployeeID 

我改变DATEPART(Day, TimeIn/Out)到CONVERT(DATE, TimeIn/Out)处理不同月/年。根据你的说法，这可能不是必要的，但是性能总体上与DATEPART()大致相同。

基本上，此处的方法是获取每个employeeID不同天数的列表，无论该日期是来自TimeIn还是TimeOut列。 UNION完美适用于此，因为它可以组合两列，并从结果集中删除重复结果，这为结果集留下了明确的days/employeeID列表。然后，这只是计算每个员工ID的行数。