Convert 7 Digit Julian Date To A Calendar Date MMDDYYYY in a Access Query

I have a 7 digit number stored in a text field that is actually a Julian Date that needs to be converted to a calendar date, format mmddyyyy in a Microsoft Access Query.
rjordanbotsAsked:
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Rey Obrero (Capricorn1)Commented:
<The date should read 11012032 >

in the function above

change this line

        CJulian2Date = Format(DateSerial(YYYY, 1, JulDay), "m/d/yyyy")

with

        CJulian2Date = Format(DateSerial(YYYY, 1, JulDay), "mmddyyyy")
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englanddgCommented:
Try using DateSerial(Left([FieldName],2),1,Right([FieldName],3))
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Rey Obrero (Capricorn1)Commented:
are you sure that what you have are Julian dates?
post sample values.
you can use this code for conversion

Function CJulian2Date (JulDay As Integer, Optional YYYY)
    If IsMissing(YYYY) Then YYYY = Year(Date)
    If Not IsNumeric(YYYY) Or YYYY \ 1 <> YYYY Or YYYY < 100 Or YYYY _
      > 9999 Then Exit Function
    If JulDay > 0 And JulDay < 366 Or JulDay = 366 And _
      YYYY Mod 4 = 0 And YYYY Mod 100 <> 0 Or YYYY Mod 400 = 0 Then _
        CJulian2Date = Format(DateSerial(YYYY, 1, JulDay), "m/d/yyyy")
End Function


'    JulDay: The ordinal day of a year. Between 1 and 365 for all
'            years, or between 1 and 366 for leap years.
'
'    YYYY: A three or four digit integer for a year that is within the
'          range of valid Microsoft Access dates. If YYYY is omitted,
'          then YYYY is assumed to be the year of the current system
'          date.
'

codes taken from this link
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KRUNAL TAILORFreelance Mobile DeveloperCommented:
Hi rjordanbots,

select DateSerial(Left([JulianField],2),1,Right([JulianField],3))
from table
 
You can also write a function in your database. like this_

Function JulianToDate(JulianDate As String) As Date

  Dim astrDatePart() As String
  
  astrDatePart = Split(JulianDate, ".")
  
  JulianToDate = DateSerial(CInt(astrDatePart(0)), 1, CInt(astrDatePart(1)))

End Function

Open in new window



Then in the query we can use the function directly.
 
select JulianToDate(table.[Myfield])
from table

More Refrence here: http://support.microsoft.com/kb/162745

Thanks & Regards,
Krunal T. Tailor
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rjordanbotsAuthor Commented:
Can you tell me what Julian Date 2032306 should convert to ?

I believe it should be 11012032
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KRUNAL TAILORFreelance Mobile DeveloperCommented:
Hi rjordanbots,

Please check over here http://aa.usno.navy.mil/data/docs/JulianDate.php

There are so many online converter available like_
1) http://www.onlineconversion.com/julian_date.htm
2) http://www.aavso.org/jd-calculator

Thanks & Regards,
Krunal T. Tailor
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Rey Obrero (Capricorn1)Commented:
if that (2032306) is a julian date the conversion is 7/22/2306 using the function above
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rjordanbotsAuthor Commented:
This number is coming off of a Ibm I-Series MainFrame. I believe the first four digits is the year then the last three is the number of days. Can this be converted in Access to an actual date ?
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englanddgCommented:
If you have Julien dates...@Krunal has your solution.

That code is solid.
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englanddgCommented:
Oh, data dump?

Yes.  Use the right and left functions.
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Rey Obrero (Capricorn1)Commented:
believe the first four digits is the year then the last three is the number of days. Can this be converted in Access to an actual date ? Yes

using the function

CJulian2Date(306,2032)

the value will be  11/1/2032
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KRUNAL TAILORFreelance Mobile DeveloperCommented:
Yes,  Rey Obrero is correct.

Please check here more ref: http://support.microsoft.com/kb/162745
Something like this:

CJulian2Date(32, 1996)
Note that your system's date format for 2/1/96 is returned. If your system's date is in 1996 then CJulian2Date(32) will also return 2/1/96.

Thanks & Regards,
Krunal T. Tailor
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rjordanbotsAuthor Commented:
The first four digits is the year 2032, the last 3 are the # of days, which is 306, need to convert the last 3 digits, which appears to be the julian day to an actual date

The date should read 11012032
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rjordanbotsAuthor Commented:
Thanks for your help, all of this information was very helpful.
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