To find the verification digit for a given value.

Aravindh_Navision

Hai Friends,

I need to find the verification digit to the following value: 3070841326319000361243033388633240620117 (Length - 39 char)

Logic for finding the Verification Digit:

1. Sum odd positions [3, 7, 8, 1, 2, 3, 9, 0, 3, 1, 4, 0, 3, 8, 6, 3, 4, 6, 0, 1]. (SumOdd = 72)
2. Multiply SumOdd * 3. (Mult = 72 * 3 = 216)
3. Sum even positions [0, 0, 4, 3, 6, 1, 0, 0, 6, 2, 3, 3, 3, 8, 3, 2, 0, 2, 1]. (SumEven = 47)
4. Sum = Mult + SumEven, (Sum = 216 + 47 = 263)
5. Verification digit 7. (We need to add the integer value 7 to 263 so the nearest higher value which should divisible by 10 and the remainder remains 0.)

To find the lowest number (verification digit):

VeriDigit = 10 - (Sum - (int(Sum/10) * 10))
VeriDigit = 10 - (263 -(int(263/10) * 10))
VeriDigit = 10 - (263 - (26 * 10))
VeriDigit = 10 - (263 - 260) = 10 - 3
VeriDigit = 7 (Answer)

Can anyone please help me in solving this in NAV?

Thanks in advance.

mohana_cse06

Create a function and pass the Sum value and get return value as

EXIT(Sum - (int(Sum/10) * 10))

Aravindh_Navision

Thanks Mohana. How to operate with the given 39 char length value? This contains 2 types of logic. How to proceed with the first one?

mohana_cse06

from where do you get that value? based on that you can calculate the first 4 steps..right?

Aravindh_Navision

Actually I am getting Ptxt_String value from the combination of few fields. Could you please correct where I am missing out the logic in the below lines of coding?

Ptxt_String = 3070841326319000361243033388633240620117 (39 char)
Ltxt_resultstring, Ltxt_resultstringO, Ptxt_String are text varialbles and others are integer variables

CLEAR(LInt_total);
CLEAR(LInt_totalO);
CLEAR(LInt_TempInt);
CLEAR(LInt_TempIntO);

FOR a := 1 TO STRLEN(Ptxt_String) DO BEGIN
  EVALUATE(LInt_TempInt,COPYSTR(Ptxt_String,a,1));
  EVALUATE(LInt_TempIntO,COPYSTR(Ptxt_String,a,1));
  IF (a MOD 2) <> 0 THEN BEGIN
    LInt_TempInt *= 3;
    Ltxt_resultstring := Ltxt_resultstring + FORMAT(LInt_TempInt);
  END ELSE IF (a MOD 2) = 0 THEN BEGIN
    LInt_TempIntO *= 1;
    Ltxt_resultstringO := Ltxt_resultstringO + FORMAT(LInt_TempIntO);
  END;
END;

EVALUATE(LInt_TempInt,COPYSTR(Ptxt_String,STRLEN(Ptxt_String),1));
EVALUATE(LInt_TempIntO,COPYSTR(Ptxt_String,STRLEN(Ptxt_String),1));
Ltxt_resultstring := Ltxt_resultstring + FORMAT(LInt_TempInt);
Ltxt_resultstringO := Ltxt_resultstringO + FORMAT(LInt_TempIntO);

FOR a := 1 TO STRLEN(Ltxt_resultstring) DO BEGIN
  EVALUATE(LInt_TempInt,COPYSTR(Ltxt_resultstring,a,1));
  LInt_total += LInt_TempInt;
END;

FOR a := 1 TO STRLEN(Ltxt_resultstringO) DO BEGIN
  EVALUATE(LInt_TempIntO,COPYSTR(Ltxt_resultstringO,a,1));
  LInt_totalO += LInt_TempIntO;
END;

LInt_FinalTotal := LInt_total + LInt_totalO;

MBerger

message('%1',strchecksum('307084132631900036124303338863324062011',
                         '313131313131313131313131313131313131313')) ;

this returns your '7'

mohana_cse06

Aravindh_Navision wrote:

I need to find the verification digit to the following value: 3070841326319000361243033388633240620117 (Length - 39 char)

STRLEN is 40 not 39 in above example

Aravindh_Navision wrote:

3. Sum even positions [0, 0, 4, 3, 6, 1, 0, 0, 6, 2, 3, 3, 3, 8, 3, 2, 0, 2, 1]. (SumEven = 47)

[0, 0, 4, 3, 6, 1, 0, 0, 6, 2, 3, 3, 3, 8, 3, 2, 0, 2, 1,7]
Here Sum is 54

Aravindh_Navision wrote:

4. Sum = Mult + SumEven, (Sum = 216 + 47 = 263)

Here Sum = 216 + 54 = 270
[/quote]

Try this code

Ptxt_String := '3070841326319000361243033388633240620117';
CLEAR(LInt_total);
CLEAR(LInt_totalO);
CLEAR(LInt_TempInt);
CLEAR(LInt_TempIntO);

FOR a := 1 TO STRLEN(Ptxt_String) DO BEGIN
  EVALUATE(LInt_TempInt,COPYSTR(Ptxt_String,a,1));
  EVALUATE(LInt_TempIntO,COPYSTR(Ptxt_String,a,1));
  IF (a MOD 2) <> 0 THEN BEGIN
    LInt_TempInt *= 3;
    Ltxt_resultstring := Ltxt_resultstring + LInt_TempInt;
  END ELSE 
  IF (a MOD 2) = 0 THEN BEGIN
    LInt_TempIntO *= 1;
    Ltxt_resultstringO := Ltxt_resultstringO + LInt_TempIntO;
  END;
END;
Message('%1',Ltxt_resultstring);
Message('%1',Ltxt_resultstringO);

LInt_FinalTotal := Ltxt_resultstringO + Ltxt_resultstring;
Message('%1',LInt_FinalTotal);

Where Ltxt_resultstring and Ltxt_resultstringO also integers..

Aravindh_Navision

Mohana,

Sorry I missed to delete the last character in the given value. Yeah I will try with your code and let know about the result. Thanks you very much for your effort.

MBerger,

STRCHECKSUM function worked for me. Thank you very much. Let me try with Mohana's modified code too.


Thanks.