ISO 8601 date to seconds

[Bob Sneidar]

Sorry about the extra lines in the last post. Not sure what caused that. Here is the formatDate() function as well. 

Bob S

function formatDate theDate, theFormat
   /*
   Accepts any valid date for the first parameter. If not a valid date, it simply returns
   what was passed. Second parameter can be any of the following:
   sql date: date in the yyyy-mm-dd format
   short date, abbreviated date, internet date, long date: LC versions of the same
   julian date: Julian number based on (I believe) Jacques formula)
   */
   
   put the itemdelimiter into theOldDelim
   set the itemdelimiter to "-"
   
   if the length of item 1 of theDate = 4 and \
         the number of items of theDate = 3 and \
         item 1 of theDate is a number and \
         item 2 of theDate is a number and \
         item 3 of theDate is a number then
      put item 2 of theDate & "/" & \
            item 3 of theDate & "/" & \
            item 1 of theDate into theDate
   end if
   
   convert theDate to dateitems
   set the itemdelimiter to theOldDelim
   
   switch theFormat
      case "sql date"
         put item 1 of theDate & "-" & \
               format("%02d",item 2 of theDate) & "-" & \
               format("%02d",item 3 of theDate) into theDate
         break
      case "short date"
         convert theDate from dateitems to short date
         break
      case "abbreviated date"
         convert theDate from dateitems to abbreviated date
         break
      case "internet date"
         convert theDate from dateitems to internet date
         break
      case "long date"
         convert theDate from dateitems to long date
         break
      case "julian date"
         put the date into theDate
           convert theDate to dateItems
           if  ((item 2 of theDate = 1) or (item 2 of theDate = 2)) then
                 put 1 into theDay
           else
                 put 0 into theDay
           end if
           put item 1 of theDate + 4800 - theDay into theYear
           put item 2 of theDate + (12 * theDay) - 3 into theMonth
           put item 3 of theDate + \
                       ((153 * theMonth + 2) div 5) + \
                       (365 * theYear) + \
                       (theYear div 4) - \
                       (theYear div 100) + \
                       (theYear div 400) - \
                       32045 into theDate
         break
   end switch
   
   return theDate 
end formatDate


> On Jan 29, 2015, at 08:05 , Bob Sneidar <bobsneidar at iotecdigital.com> wrote:
> 
> Not that this solves your particular problem but some might be interested in this function. I will see if I can include this scenario in my function at some point.
> 
> Bob S