Julian Date To Calendar Date

[yeewaitu]

Does anyone know how to use impromptu to convert julian date (eg: 103118) to calendar date (28/04/03) using the basic date functions in impromptu?

Please help. Thanks!

 

[GJParker]

This will convert the example you gave but it seems a very strange start date

add-days (1720-12-29 , 103118)

or

datetime-to-date(days-from-1900-to-datetime(103118,65380))

HTH

Gary

Gary Parker
MIS Data Analyst
Manchester, England

 

[yeewaitu]

Can I ask why is the start date very strange?

Can this formula handle leap years?

 

[GJParker]

In my experience most db's tend to set there dates form the start of a year that's all.

You'll need to test the formulas to check leap years.



Gary Parker
MIS Data Analyst
Manchester, England

 

[yeewaitu]

What does 65380 mean?

 

[drlex]

Sorry to butt in, but I think yeewaitu's example is incorrect; calendar date 28/04/(20)03 is 2452758 in Julian days. I suspect this is what Gary means when he mentions 'strange'.

On this basis, Gary's solution would be

datetime-to-date(days-from-1900-to-datetime(2452758-2415020))

2415020 is 31st December 1899 in Julian


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[drlex]

This might be a more precise method, although lacking Gary's simplicity:

Let JD = Julian calendar date and day, month and year be the required Calendar date.

Then, for calendar to Julian:
d = INT((4*(JD + 32082)+3)/1461)
e = (JD + 32082) - INT((1461*d)/4)
m = INT((5*e+2)/153)

day = e - INT((153*m+2)/5) + 1
month = m + 3 - 12*INT(m/10)
year = d - 4800 + INT(m/10)

To revert from day, month, year to Julian:
a = INT(14-month)/12
y = year+4800-a
m = month + 12*a - 3

JD = day + INT((153*m+2)/5) + y*365 + INT(y/4) - 32083

(INT is integer and so you might prefer to use INTEGER-DIVIDE(x,y) if available)


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