Date To Julian Date Trouble

Legacy Poster

• 19. August 2013 22:09

My code was inspired from this example found on the web.

Function Date2Julian(vDate) 

 

    Date2Julian = CLng(Format(Year(vDate), "0000") _

                  + Format(DateDiff("d", CDate("01/01/" _

                  + Format(Year(vDate), "0000")), vDate) _

                  + 1, "000"))

    

End Function

 

This is my latest version in my many attempts to get this to work.

 

Data for label comes from XML

Shared/name = “IOLOT2”

Sample data from IOLOT2 = “EXP1602”

 

Originally use VBScript command “Format” got error,
“Unrecognized”, not FormatNumber.

 

Code that will not work

 

dim stryear

dim strMonth

dim vDate

dim strBatch

 

strBatch = IOLOT2

      

strBatch =  Mid(strBatch,4, LEN(strBatch) - 3)

 

strYear  = Left(strBatch ,2)

 

strMonth= Right(strBatch,2)

 

 vDate=strMonth & "/01/" & strYear

 

Value = CLng(FormatNumber(Year(vDate), "0000") &  FormatNumber(DateDiff("d", CDate("01/01/" & FormatNumber(Year(vDate), "0000")), vDate) & 1, "000"))

 

 

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  Legacy Poster

  ★ BarTender Hero ★

  • 20. August 2013 16:50

  I figured out the problem.

   

  I had used the script function "Format: many times, but it is unsupported by the Bartender script interpreter, and I was not familiar with "FormatNumber", so when I was getting an error from the function "CDate", I incorrectly thought that it was also unsupported . AS it turns out the FormatNumber function was adding a comma to the four digit year, and that was breaking the CDate function.

   

  Here is the code that works.

   

  dim stryear
  dim strMonth
  dim vDate
  dim strBatch
  strBatch = IOLOT2
  strBatch =  Mid(strBatch,4, LEN(strBatch) - 3)

   

      strYear  = Left(strBatch ,2)
      strMonth= Right(strBatch,2)
      vDate=strMonth & "/01/" & strYear

   

  Value = CLng(FormatNumber(Year(vDate), "0000") _
                    + FormatNumber(DateDiff("d", CDate("01/01/" _
                    + FormatNumber(Year(vDate), "0000",,,False)), vDate) _
                    + 1, "000"))

   

  
   

   

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