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shekhar_analyti
Specialist
Specialist

how to create a transformation subroutine only for one column of table for security purpose ?

Hi All ,

I have requirement in two parts :

a) How to create a  transformation subroutine only for one column of table for security purpose for first 5 numbers/letters ?
b) Transformation logic changes day wise .

  

first_namelast_nameCREDIT CARDNbrDate_of_Use
KerrillStenningswitch490349039171156022-01-2017

So credit card number should be twiked to something like this

First number from 4903490391711560 is

Logic based on Day 7th since 22/02/2017


4 + 7 (because 22 jan 2017 was 7th day of that week) = 11

9 + 1 (1st month of year) = 10
0 + (11+10) = 21
3  * 16 (Total length of Nbr) = 48

4 * 5 ( even 5 number is even then multiply by next upper odd number , if odd then add 6) = 20


Hence

4903490391711560 changes to ---------> 111021482090391711560


Logic based on Day 6th


4 + (6*2)  (because 21 jan 2017 was 6th day of that week) = 16

9 + 1 (1st month of year) = 10
0 + (11+10) = 21
3  * 16 (Total length of Nbr) = 48

4 * 5 ( even 5 number is even then multiply by next upper odd number , if odd then add 6) = 20


4903490391711560 changes to ---------> 161021482090391711560


Similarly there are different logic for all seven days of the week @


Thanks & Regards

Shekar

20 Replies
shekhar_analyti
Specialist
Specialist
Author

Sunny Bhai .. Please help .. stalwar1

Peter_Cammaert
Partner - Champion III
Partner - Champion III

A few introductory questions:

  • What is the role of date 22/2/2017 in this story? See bold black line.
  • Which date is driving the algorithm selection? Today() ? The day of the current reload? Some other date?
  • You describe 2 algorithms that only differ in the way they translate the first digit. Where are the 5 others?

Peter

shekhar_analyti
Specialist
Specialist
Author

Hi Peter ,

Thanks for reply .

Date 22/2/2017 refers to date of swipe of credit card for user . And actually it drives every thing of algorithm .

If the date of use falls on first day of week i.e Monday then there is a set of logic , if its on 2nd day of week i.e Tuesday then there is different logic . It is just that i have highlighted logic for 7th day because 22 jan 2017 was sunday .

Other 5 algorithm i will be taking care , after going through solution for above two days

Frank_Hartmann
Master II
Master II

Logic based on Day 6th


4 + (6*2)  (because 21 jan 2017 was 6th day of that week) = 16      //Why multiplying  by 2??

9 + 1 (1st month of year) = 10
0 + (11+10) = 21                                                                         //If above is correct why not 16 +10???
3  * 16 (Total length of Nbr) = 48

4 * 5 ( even 5 number is even then multiply by next upper odd number , if odd then add 6) = 20


Im not getting your logic behind this!

shekhar_analyti
Specialist
Specialist
Author

4 + (6*2)  (because 21 jan 2017 was 6th day of that week) = 16      //Why multiplying  by 2??

4 + (6th day * 2) its a fixed number , 2 is fixed number  

9 + 1 (1st month of year) = 10

0 + (11+10) = 21  .. year your are correct it should be 16+10 , agreed my mistake    

Frank_Hartmann
Master II
Master II

so if its the fifth day it should be multiplyed by 3 and so on?

because in Logic based on Day 7th you did not multiply by 2

shekhar_analyti
Specialist
Specialist
Author

Actually multiplication by 2 is only for day 6 , not for day 7

shekhar_analyti
Specialist
Specialist
Author

for 5 day the logic is different .. completely different

Frank_Hartmann
Master II
Master II

then i guess we need to know the complete logic