Generating Luhn Checksums

Question

There are lots of implementations for validating Luhn checksums but very few for generating them. I\'ve come across this one however in my tests it has revealed to be buggy

Answer 1

BAD

I literally cannot believe how many crummy implementations there are out there.

IDAutomation has a .NET assembly with a MOD10() function to create but it just doesn't seem to work. In Reflector the code is way too long for what it's supposed to be doing anyway.

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BAD

This mess of a page which is actually currently linked to from Wikipedia(!) for Javascript has several verification implementations that don't even return the same value when I call each one.

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GOOD

The page linked to from Wikipedia's Luhn page has a Javascript encoder which seems to work :

// Javascript
String.prototype.luhnGet = function()
{
    var luhnArr = [[0,1,2,3,4,5,6,7,8,9],[0,2,4,6,8,1,3,5,7,9]], sum = 0;
    this.replace(/\D+/g,"").replace(/[\d]/g, function(c, p, o){
        sum += luhnArr[ (o.length-p)&1 ][ parseInt(c,10) ]
    });
    return this + ((10 - sum%10)%10);
};

alert("54511187504546384725".luhnGet());​

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GOOD

This very useful EE4253 page verifies the check-digit and also shows the full calculation and explanation.

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GOOD

I needed C# code and ended up using this code project code:

// C#
public static int GetMod10Digit(string data)
        {
            int sum = 0;
            bool odd = true;
            for (int i = data.Length - 1; i >= 0; i--)
            {
                if (odd == true)
                {
                    int tSum = Convert.ToInt32(data[i].ToString()) * 2;
                    if (tSum >= 10)
                    {
                        string tData = tSum.ToString();
                        tSum = Convert.ToInt32(tData[0].ToString()) + Convert.ToInt32(tData[1].ToString());
                    }
                    sum += tSum;
                }
                else
                    sum += Convert.ToInt32(data[i].ToString());
                odd = !odd;
            }

            int result = (((sum / 10) + 1) * 10) - sum;
            return result % 10;
        }

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GOOD

This validation code in C# seems to work, if a little unwieldy. I just used it to check the above was correct.