如何将 Excel 序列日期号转换为 .NET DateTime？

Question

How do I convert from excel serial date to a .NET date time?

For example 39938 is 05/05/2009.

Answer 1

I find it simpler using FromOADate method , for example:

DateTime dt = DateTime.FromOADate(39938);

Using this code dt is "05/05/2009".

Answer 2

Where 39938 is the number of days since 1/1/1900?

In that case, use the framework library function DateTime.FromOADate() .
This function encapsulates all the specifics, and does bounds checking.

For its historical value, here is a possible implementation:

(C#)

public static DateTime FromExcelSerialDate(int SerialDate)
{
    if (SerialDate > 59) SerialDate -= 1; //Excel/Lotus 2/29/1900 bug   
    return new DateTime(1899, 12, 31).AddDays(SerialDate);
}

VB

Public Shared Function FromExcelSerialDate(ByVal SerialDate As Integer) As DateTime
    If SerialDate > 59 Then SerialDate -= 1 ' Excel/Lotus 2/29/1900 bug
    Return New DateTime(1899, 12, 31).AddDays(SerialDate)
End Function

---

[Update]:
Hmm... A quick test of that shows it's actually two days off. Not sure where the difference is.

Okay: problem fixed now. See the comments for details.

Answer 3

For 39938 do this: 39938 * 864000000000 + 599264352000000000

864000000000 represents number of ticks in a day 599264352000000000 represents number of ticks from the year 0001 to the year 1900

Answer 4

void ExcelSerialDateToDMY(int nSerialDate, int &nDay, 
                          int &nMonth, int &nYear)
{
    // Excel/Lotus 123 have a bug with 29-02-1900. 1900 is not a
    // leap year, but Excel/Lotus 123 think it is...
    if (nSerialDate == 60)
    {
        nDay    = 29;
        nMonth    = 2;
        nYear    = 1900;

        return;
    }
    else if (nSerialDate < 60)
    {
        // Because of the 29-02-1900 bug, any serial date 
        // under 60 is one off... Compensate.
        nSerialDate++;
    }

    // Modified Julian to DMY calculation with an addition of 2415019
    int l = nSerialDate + 68569 + 2415019;
    int n = int(( 4 * l ) / 146097);
            l = l - int(( 146097 * n + 3 ) / 4);
    int i = int(( 4000 * ( l + 1 ) ) / 1461001);
        l = l - int(( 1461 * i ) / 4) + 31;
    int j = int(( 80 * l ) / 2447);
     nDay = l - int(( 2447 * j ) / 80);
        l = int(j / 11);
        nMonth = j + 2 - ( 12 * l );
    nYear = 100 * ( n - 49 ) + i + l;
}

Cut and Paste of someone elses talents...

Ian Brown