C#: Adding working days from a cetain date

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不知归路
不知归路 2021-02-14 18:10

I have trouble doing this. I\'m creating a method that add working days on a specific date. for example, I want to add 3 working days to sept 15, 2010 (Wednesday), the method wo

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  •  时光说笑
    2021-02-14 18:58

    If you don't need to consider holidays, I would suggest you do something like this:

    public static DateTime AddWorkingDays(DateTime specificDate,
                                          int workingDaysToAdd)
    {
        int completeWeeks = workingDaysToAdd / 5;
        DateTime date = specificDate.AddDays(completeWeeks * 7);
        workingDaysToAdd = workingDaysToAdd % 5;
        for (int i = 0; i < workingDaysToAdd; i++)
        {
            date = date.AddDays(1);
            while (!IsWeekDay(date))
            {
                date = date.AddDays(1);
            }
        }
        return date;
    }
    
    private static bool IsWeekDay(DateTime date)
    {
        DayOfWeek day = date.DayOfWeek;
        return day != DayOfWeek.Saturday && day != DayOfWeek.Sunday;
    }
    

    It's inefficient, but easy to understand. For an efficient version, you'd work out the number of complete weeks to add as before, but then have a mapping from any "current day of week" and "working days left to add" to "number of actual days to add". Then you could just work out the total number of days to add, and do it in one call.

    EDIT: In terms of the level of inefficiency... it's really not very bad. It'll only perform manual "is this a weekend" checks for up to 4 days, which isn't too bad. In particular, despite igor's (current at the time of posting) claims, it's rather faster than his approach, flawed benchmarks notwithstanding ;)

    Note that it may not handle negative inputs yet - I haven't checked.

    One of the reasons behind the approach I'm using is that it doesn't rely on either me or the code reader knowing what the values in the DayOfWeek enum are. I don't care whether it's 0-6, 1-7, Monday-Sunday, Saturday-Friday... or even if there are completely bizarre values. I only compare for equality, which makes the code more "obviously correct".

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