Tomorrow's date

• 03/26/2019 at 1:00 PM #94722
  dburgh

  Is there a function or a way to get tomorrow’s date?

  Ideally, I’d like to do something like this..

  today + 3 and then check if that date is part of the current month or not.

  Thanks,

  David

  

  dburgh

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  03/26/2019 at 1:52 PM #94725
  robertogozzi

  You can find the last day of the current month and check if it’s beyond:

  DayMax = 31 If Month = 4 or Month = 6 or Month = 9 Then DayMax = 30 Endif If Month = 2 then DayMax = 28 If year mod 4 = 0 then If year mod 100 = 0 then If year mod 400 = 0 then DayMax = 29 Endif Else DayMax = 29 Endif Endif Endif

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  DayMax = 31 If Month = 4 or Month = 6 or Month = 9 Then    DayMax = 30 Endif If Month = 2 then    DayMax = 28    If year mod 4 = 0 then       If year mod 100 = 0 then          If year mod 400 = 0 then             DayMax = 29          Endif       Else          DayMax = 29       Endif    Endif Endif

   

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  03/26/2019 at 1:57 PM #94726
  dburgh

  This is beautiful! How do I grab currentdate+3day’s date to compare to the max?

  

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  03/26/2019 at 1:58 PM #94728
  dburgh

  But I am not sure if this helps.. as I want a count of business days remaining in the month.

  

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  03/26/2019 at 2:13 PM #94729
  robertogozzi

  Current Year is YEAR

  Current Month is MONTH

  Current Day is DAY.

  If you need to extract the day from a date, say 20190326, then you’ll have to divide it by 1000000 (it’s a common number for Prt), then round it to the lowest integer, then multiply it by 1000000, then compute the difference between the date and the number you have got:

  x = date y = round((x / 1000000) - 0.5) z = y * 1000000 d = date - z

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  x = date y = round((x / 1000000) - 0.5) z = y * 1000000 d = date - z

  d is the DAY number.

  

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  03/26/2019 at 2:19 PM #94730
  robertogozzi

  This is a link to my code to compute business days in a month https://www.prorealcode.com/topic/trading-in-giorni-specifici/#post-64933

  “Giornoferiale” means business day.

  

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  03/26/2019 at 2:20 PM #94731
  dburgh

  IF OpenMonth <> OpenMonth[1] THEN                                            //Se il mese della barra attuale è diverso da quello della barra precedente significa che siamo alla prima barra del mese…

    BusinessDay = 0                                                         //… per cui dobbiamo ricominciare da 0 a contare i giorni lavorativi

  ENDIF

  IF OpenDayOfWeek >= 1 AND OpenDayOfWeek <= 5 AND IntraDayBarIndex = 0 THEN   //Se siamo tra lunedì (1) e venerdì (5) ed è la prima barra del giorno…

     BusinessDay = BusinessDay + 1                                         //… significa che è iniziato un nuovo giorno lavorativo.

  ENDIF

  But how can I create a function that returns 3 when it checks today’s date 20190326?

  Using the above would give me the current businessday count for current date. Not sure how to get businessday count for last day of the month. if I had that value can do simple subtraction.

  

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  03/26/2019 at 2:55 PM #94733
  robertogozzi

  Sorry, my previous code was a divion (then multiplication) by 100, not 1000000!

  Just modify my previous DATE code:

  Get the DAY

  x = date //x = 20190326 y = round((x / 100) - 0.5) //y = 201903.00 z = y * 100 //z = 20190300 d = x - z //d = 26

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  x = date                     //x = 20190326 y = round((x / 100) - 0.5)   //y = 201903.00 z = y * 100                  //z = 20190300 d = x - z                    //d = 26

  into Get MONTH:

  Get the MONTH

  x = date //x = 20190326 y = round((x / 100) - 0.5) //y = 201903.00 z = round((y / 100) - 0.5) //z = 2019.0000 m = y - z //m = 3

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  x = date                     //x = 20190326 y = round((x / 100) - 0.5)   //y = 201903.00 z = round((y / 100) - 0.5)   //z = 2019.0000 m = y - z                    //m = 3

   

  

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  03/26/2019 at 3:40 PM #94736
  dburgh

  It is not making sense. How can I combine what you’re providing to return the number of business days left in a month? Can you show me a function that takes today’s date 20160326 and returns 3. The same function should take tomorrow’s date and return 2.

   

   

  

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  03/26/2019 at 4:10 PM #94737
  Nicolas

  day of the week number can be retrieved with: (Replace myYear,myMonth and myDay with YYYY, MM, DD number format. In this example it basically returns the day of the week of the current day.), I think that we could associate the snippets from Roberto to this one to know the last business day.

  //find the day number with a date myYear=Year myMonth=Month myDay=Day if myMonth >= 3 then D = (((23*myMonth)/9) + myDay + 4 + myYear + (myYear/4) - (myYear/100) + (myYear/400) - 2) mod 7 else z = myYear - 1 D = (((23*myMonth)/9) + myDay + 4 + myYear + (z/4) - (z/100) + (z/400) ) mod 7 endif return d

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  //find the day number with a date myYear=Year myMonth=Month myDay=Day   if myMonth >= 3 then D = (((23*myMonth)/9) + myDay + 4 + myYear + (myYear/4) - (myYear/100) + (myYear/400) - 2) mod 7 else z = myYear - 1 D = (((23*myMonth)/9) + myDay + 4 + myYear + (z/4) - (z/100) + (z/400) ) mod 7 endif     return d

   

  

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  03/26/2019 at 4:15 PM #94738
  robertogozzi

  There’s no built-in function, by combining the above codes (extracting from dates) and business day math you can achieve that, but it takes a loop, since being 3 does not mean there have been 3 busniss days. PRT does not support easy date math (like eXcel does).

  You’ll have to start a loop from the day you want, then go backwords counting only those days whose dayofweek is between 1 and 5 to find out how many business day there were BEFORE that day.

  It’s not as straightfoward!

  

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  03/26/2019 at 4:49 PM #94739
  dburgh

  I cannot understand how to replace your ‘OpenDayOfWeek’ in the businessday logic.

   

  Is there a way to get the day of the week from the date?

   

  My temporary idea in pseudo code

  x = date

  DayMax = getDayMax(x)

  businessdayleft = 0

   

  for count = 1 to DayMax

  new date = x + count

  if dayNumber(newdate) > DayMax then break

  if getDayOfWeek(newdate) is business then increment bussinessdaysleft by 1

   

   

  // basically I need a dayNumber and dayOfWeek function unless I’m missing something?

  

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  03/27/2019 at 12:36 PM #94772
  dburgh

  we can use Nicolas’s solution for dayNumber, but am still unsure how to extract dayOfWeek

  

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  03/27/2019 at 1:31 PM #94776
  robertogozzi

  I’m working on it.

  

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  03/27/2019 at 4:43 PM #94795
  robertogozzi

  OpenDayOfWeek returns:

  1=Monday, … 5=Friday, so if any OpenDayOfWeek is within the range 1-5 it’s a business day (unless there’s some local holiday), so you can detect how many business days there are in each month by tallying from day 1 to the last day and skipping those ourside that range.

  Anyway, I attach 4 functions (they are actually indicators, but the behaviour is the same as in other languages, they may require parameters as their input and return some data):

  • DayMax                 (input: Month,Year                                                               returns: MaxDay)
  • IsLeapYear          (input: Year                                                                             returns: 1=leap year, 0=no leap year)
  • UnpackDate        (input: Date formatted as YYYYMMDD, as PRT does    returns: Day,Month,Year)
  • GetDayOfWeek  (input: Date formatted as YYYYMMDD, as PRT does    returns: 0=Sunday,1=Monday,….,6=Saturday)

  I attach both .ITF files to be imported into ProBuilder as well as .TXT files:

  DayMax

  //input: MONTH,YEAR //requires indicator ISLEAPYEAR // MaxDay = 31 If MyMonth = 4 or MyMonth = 6 or MyMonth = 9 or MyMonth = 11 Then MaxDay = 30 Endif If MyMonth = 2 then x = CALL "IsLeapYear"[MyYear] MaxDay = 28 + x Endif RETURN MaxDay

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  //input:  MONTH,YEAR //requires indicator ISLEAPYEAR // MaxDay = 31 If MyMonth = 4 or MyMonth = 6 or MyMonth = 9 or MyMonth = 11 Then    MaxDay = 30 Endif If MyMonth = 2 then    x = CALL "IsLeapYear"[MyYear]    MaxDay = 28 + x Endif RETURN MaxDay

  IsLeapYear

  //input: YEAR LeapYear = 0 If MyYear mod 4 = 0 then If MyYear mod 100 = 0 then If MyYear mod 400 = 0 then LeapYear = 1 Endif Else LeapYear = 1 Endif Endif RETURN LeapYear

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  //input: YEAR LeapYear = 0 If MyYear mod 4 = 0 then    If MyYear mod 100 = 0 then       If MyYear mod 400 = 0 then          LeapYear = 1       Endif    Else       LeapYear = 1    Endif Endif RETURN LeapYear

  UnpackDate

  //input: DATE d = 0 m = 0 y = 0 // example: x = MyDate //x = 20190326 w = round((x / 100) - 0.5) //w = 201903.00 z = w * 100 //z = 20190300 d = x - z //d = 26 // y = round((w / 100) - 0.5) //z = 2019.0000 m = w - (y * 100) //m = 3 // RETURN d,m,y

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  //input:   DATE d = 0 m = 0 y = 0 //                              example: x = MyDate                   //x = 20190326 w = round((x / 100) - 0.5)   //w = 201903.00 z = w * 100                  //z = 20190300 d = x - z                    //d = 26 // y = round((w / 100) - 0.5)   //z = 2019.0000 m = w - (y * 100)            //m = 3 // RETURN d,m,y

  GetDayOfWeek

  //input: DATE //requires indicators UNPACKDATE and ISLEAPYEAR // // https://www.wikihow.it/Calcolare-il-Giorno-della-Settimana (method 1) // MyDay, MyMonth, MyYear = CALL "UnpackDate"[MyDate] IF MyMonth = 1 OR MyMonth = 10 THEN Mvalue = 0 ELSIF MyMonth = 4 OR MyMonth = 7 THEN Mvalue = 6 ELSIF MyMonth = 9 OR MyMonth = 12 THEN Mvalue = 5 ELSIF MyMonth = 5 THEN Mvalue = 1 ELSIF MyMonth = 6 THEN Mvalue = 4 ELSIF MyMonth = 8 THEN Mvalue = 2 ENDIF d = MyDay + Mvalue WHILE d > 6 d = d - 7 WEND y = MyYear MOD 100 Century = MyYear - y Century = (Century MOD 400) / 100 IF Century = 1 THEN Century = 5 ELSIF Century = 2 THEN Century = 3 ELSIF Century = 3 THEN Century = 1 ENDIF z = y WHILE y > 27 y = y - 28 WEND y = y + round((z / 4) - 0.5) + Century IF MyMonth < 3 THEN x = CALL "IsLeapYear"[MyYear] y = y - x ENDIF d = d + y WHILE d > 6 d = d - 7 WEND IF d = 0 THEN d = 7 ENDIF RETURN d - 1 //0=Sunday, 1=Monday...... 6=Saturday

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  //input: DATE //requires indicators UNPACKDATE and ISLEAPYEAR // // https://www.wikihow.it/Calcolare-il-Giorno-della-Settimana  (method 1) // MyDay, MyMonth, MyYear = CALL "UnpackDate"[MyDate] IF MyMonth = 1 OR MyMonth = 10 THEN    Mvalue = 0 ELSIF MyMonth = 4 OR MyMonth = 7 THEN    Mvalue = 6 ELSIF MyMonth = 9 OR MyMonth = 12 THEN    Mvalue = 5 ELSIF MyMonth = 5 THEN    Mvalue = 1 ELSIF MyMonth = 6 THEN    Mvalue = 4 ELSIF MyMonth = 8 THEN    Mvalue = 2 ENDIF d = MyDay + Mvalue WHILE d > 6    d = d - 7 WEND y = MyYear MOD 100 Century = MyYear - y Century = (Century MOD 400) / 100 IF Century = 1 THEN    Century = 5 ELSIF Century = 2 THEN    Century = 3 ELSIF Century = 3 THEN    Century = 1 ENDIF z = y WHILE y > 27    y = y - 28 WEND y = y + round((z / 4) - 0.5) + Century IF MyMonth < 3 THEN    x = CALL "IsLeapYear"[MyYear]    y = y - x ENDIF d = d + y WHILE d > 6    d = d - 7 WEND IF d = 0 THEN    d = 7 ENDIF RETURN d - 1      //0=Sunday, 1=Monday...... 6=Saturday

  Edit:     The above GetDayOfWeek function does not work fine, it has been replaced by a new version  (see post https://www.prorealcode.com/topic/tomorrows-date/page/2/#post-95272)

   

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