Elapse days - calculation

[Former Member]

I have Month and year field. Values Year(2009) and Month(4)

I want below logic to calculate -

MTD qty % = (qty * elapse days) / no of days in month

so from the above, I want to know,

1. How can we get the no of days from the above 2 objects (year and month).

2. Elapse days are the days that are over from current date to Ist day of that month.

eg: current date - 04.03.2009 , 1st day of the month - 04.01.2009,

elapse days = 2. (you need to consider all days in the month, not only working days)

so my questions is how can we get, # of days n a month from above 2 fields and elapse days based on the above condition.

please provide your suggestions.

Thanks,

Pra

[anuj_srivastava]

Hi ,

u can use this function module to find the no of days in that particular month.

HR_E_NUM_OF_DAYS_OF_MONTH .

For calculating the elapsed days u can put the simple arithmetic logic.

Elapsed day = sy-datum - 01/month/year.

While doing the arithmetic calculation the date should be of type datum.

Regards,

Anuj

[Former Member]

Hi,

I want to use only Month and Year to get the no of days in the month.

Please provide me how to convert date(01 - constant), month, year into datum format.

thanks

[anuj_srivastava]

Hi ,

If u will give the specific month as input to the FM it will return u the no of days in that specific month.

Regarding changing the date into datum format you need to write it into this format for month (04) and year ( 2009) as 20090401.

This can be achived using a simple concatenate statement.

Concatenate year month constant into date .

now you can do arithmetic operation with it.

Regards,

Anuj

[former_member156446]

check this code which I wrote for you:

REPORT  zj_test LINE-SIZE 162.

TABLES: tfacs.
DATA: lv_num TYPE i, lv_string TYPE char40.

DATA: BEGIN OF itab.
        INCLUDE STRUCTURE tfacs.
DATA: END OF itab.

SELECT SINGLE *
  FROM tfacs
  INTO itab
  WHERE ident = 'US'
  AND  jahr = '2009'.

lv_string = itab-mon02. "for feb

lv_num = STRLEN( lv_string ).

WRITE: lv_num.

[Former Member]

Hello,

I think the sample program below does what you ask: it finds the days in the month (bit of an overkill to use function modules for that) and then computes the QTD.

Note that the internal string representation of a data variable is always YYYYMMDD, regardless of the "externalized" form (e.g. yyyy/mm/dd or dd.mm.yyyy), so this code will work regardless of your custom date format.

Regards,

Mark

REPORT  zqty_to_date.
PARAMETERS:
  p_date     TYPE dats,
  p_qty      TYPE i.

DATA:
  days      TYPE i,
  n_year    TYPE numc4,
  n_month   TYPE numc2,
  n_day     TYPE numc2,
  qtd(6)    TYPE p DECIMALS 1.


START-OF-SELECTION.
  n_year = p_date+0(4).
  n_month = p_date+4(2).
  n_day = p_date+6(2).

  PERFORM days_in_month USING n_year n_month CHANGING days.
  qtd = ( p_qty * ( n_day - 1 ) ) / days.
  WRITE: / 'Days in month:', days,
         / 'Qty  to date :', qtd.

*&---------------------------------------------------------------------*
*&      Form  days_in_month
*&---------------------------------------------------------------------*
*       text
*----------------------------------------------------------------------*
FORM days_in_month USING year month CHANGING days.
  DATA: ymod4   TYPE i,
        ymod100 TYPE i,
        ymod400 TYPE i.
  CASE month.
    WHEN 4 OR 6 OR 9 OR 11.
      days = 30.
    WHEN 2.
      ymod4 = year MOD 4.
      ymod100 = year MOD 100.
      ymod400 = year MOD 400.
      IF ( ymod4 = 0 AND ymod100 > 0 ) OR ( ymod100 = 0 AND ymod400 = 0 ).
        days = 29.
      ELSE.
        days = 28.
      ENDIF.
    WHEN OTHERS.
      days = 31.
  ENDCASE.
ENDFORM.                    "days_in_month

[Former Member]

solved