正确计算 SQL 中的工作日

Question

I have this a table with some user id events ( starting table ).

In this table there is a Event_Id = 1 that refers to a Local Holiday (it's not a working day).

I need to fill the column 'Event_1' to be able to calculate correctly the WorkingDays ( result table - below ).

• In each row where the Event_Id <> 1, the Event_1 column should have a 1 if the corresponding user as matching Event_Id = 1 between the StartDate and EndDate, else it should be 0.

• If the row in analysis as Event_id = 1 the value in Event_1 column
  should be 0 as well.

I already have a function that calculates the WorkingDays.

Regards, Elio Fernandes

Answer 1

I came up with this solution, but I would like to know if you think it's the best way to do it.

Select T1.*,
    Case
        When t2.StartDate Is not Null and t1.Event_Id = 1 then 0
        When t2.StartDate Is not Null Then 1 
        Else 0
    End As [Local_Holiday]
From 
    [plann].[Events]  as T1 left Join
    (Select StartDate
        From [plann].[Events] 
        Where Event_Id = 1 ) as T2 ON T2.StartDate Between T1.StartDate and T1.EndDate

Answer 2

No. You can move the filter to the on clause and get rid of the subquery:

Select e1.*,
       (Case When e2.StartDate Is not Null and e1.Event_Id = 1 then 0
             When e2.StartDate Is not Null Then 1 
             Else 0
        End) As [Local_Holiday]
From [plann].[Events] e left Join
     [plann].[Events] e2
     on e2.Event_Id = 1 and
        e2.StartDate Between e1.StartDate and e1.EndDate;

Answer 3

Comparing the data results between your blue table and green table, it appears that the value of Event_1 is the number of days assigned to the “Local Holiday.”

If a “Local Holiday” will only EVER equal one day and all “Local Holiday” records apply to any user whose vacation coincides with the holiday, then Gordon's solution works (and I like how he gets rid of the subquery).

However, you may need to address other considerations.

Per your data results, each user with a Local Holiday that occurs during their vacation has a separate record indicating this, which means that the existence of a holiday record per user matters. If so, include an additional join for [User Ud].

LEFT JOIN plann.Events e2 ON e2.Event_Id = 1 
    AND e2.StartDate BETWEEN e1.StartDate AND e1.EndDate 
    AND e2.[User Ud] = e1.[User Ud]

WITHOUT the [User Ud] join, if user 22 has a holiday within his vacation time, then the holiday is recognized—but his data result is inconsistent compared to the others because he does not have a record assigned to him with an Event_Id = 1.

WITH the [User Ud] join, you get consistent results, but this leads to design issues and limitations.

Suppose a “Local Holiday” pertains to all users (typical). If so, I suggest creating a separate holiday table. Accessing the table for applicable dates ensures that all holidays are recognized for all users with an applicable vacation date range.

I modified the script removing the assumption that a holiday is always one day (Canada, the UK, and other countries have two public holidays in a row). This consideration led me back to using a subquery.

SELECT
    e1.[User Ud],
    e1.Event_Id,
    e1.Event,
    e1.StartDate,
    e1.EndDate,
    CASE
        WHEN e1.Event_Id = 1
        THEN 0
        WHEN e2.[User Ud] IS NOT NULL
        THEN Number_Days
        ELSE 0
    END AS Holiday_Nmbr_Days,
    WorkingDays

FROM plann.Events e1

LEFT JOIN
    (SELECT
        [User Ud],
        DATEDIFF(dd,StartDate,EndDate) + 1 AS Number_Days,
        StartDate
    FROM plann.Events
    WHERE Event_Id = 1) AS e2 ON e2.Startdate BETWEEN e1.Startdate AND e1.EndDate
        AND e2.[User Ud] = e1.[User Ud] /*Exclude if not applicable*/