PLSQL 中的瑞士工资单的校验位

Question

I have a problem with Luhn Algorithm for Pay Slip in Swiss. I found an algorithm in C, Phyton, or Java Script but I don't know how to implement this algorithm on Oracle.

http://dnando.github.io/blog/2014/09/23/check-digit-computation-swiss-pay-slips/

http://www.hosang.ch/modulo10.aspx

This page shows how to algorithm looks like.

public static int modulo10(string nummer)
{
    // 'nummer' darf nur Ziffern zwischen 0 und 9 enthalten!

    int[] tabelle = { 0, 9, 4, 6, 8, 2, 7, 1, 3, 5 };
    int uebertrag = 0;

    foreach (char ziffer in nummer)
        uebertrag = tabelle[(uebertrag + ziffer - '0') % 10];

    return (10 - uebertrag) % 10;
}

Do You know how to implement a table in PLSQL?

Thanks in advance for your help with this algorithm.

Answer 1

You can use:

CREATE FUNCTION luhn_modulo(
  value IN VARCHAR2
) RETURN NUMBER DETERMINISTIC
IS
  offsets CONSTANT CHAR(10) := '0946827135';
  output  PLS_INTEGER := 0;
BEGIN
  FOR i IN 1 .. LENGTH( value ) LOOP
    output := SUBSTR( offsets, MOD(output + SUBSTR( value, i, 1 ), 10) + 1, 1 );
  END LOOP;
  RETURN MOD(10 - output, 10);
END;
/

Then:

BEGIN
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '1' ) );
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '2' ) );
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '3' ) );
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '4' ) );
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '1234' ) );
  DBMS_OUTPUT.PUT_LINE( luhn_modulo( '000' ) );
END;
/

Outputs:

 1 6 4 2 7 0

db<>fiddle here