Try and catch for check sum validation on input to check three parts of an input in Javascript

Question

Bouncing my head off the wall here trying to figure out a better way to handle this. I have a large input value which has three checks to check the sum of certain parts of the string in order to validate it. I'm using three try/catch blocks in one function to run the check right now and it seems to be working except for the final validation check which always seems to return true. What I'm wondering is a) is this a good method to use, b) is there a cleaner way to do this with for loop and c) why my final check is not doing anything. Any help is appreciated. I have access to jQuery and Underscore.js if that helps but I have not worked much with underscore. I made a fiddle here:

Sample Fiddle

     window.onkeyup = keyup;
    var number;

    function keyup(e) {
        number = e.target.value;
        $('#numberValue').text(number);
        // must be 10 characters long
        if (number.length !== 30) {
            return false;
        }
        number = "" + (number || "");
        // run the checksum
        var valid = false;
        try {
            var sum = (parseInt(number[0]) * 7) +
                    (parseInt(number[1]) * 3) +
                    (parseInt(number[2])) +
                    (parseInt(number[3]) * 7) +
                    (parseInt(number[4]) * 3) +
                    (parseInt(number[5])) +
                    (parseInt(number[6]) * 7) +
                    (parseInt(number[7]) * 3) +
                    (parseInt(number[8]));
            alert(((sum % 10).toFixed(0)));
            var checkDigit = ((sum % 10).toFixed(0));

            if ((number[9]) === ("" + checkDigit)) {
                alert('Our Checkdigit is valid', checkDigit);
                valid = true;
            }
        } catch (e) {
            alert('Fail for check 1!');
            valid = false;
        }
        try {
            var sum2 = (parseInt(number[13]) * 7) +
                    (parseInt(number[14]) * 3) +
                    (parseInt(number[15])) +
                    (parseInt(number[16]) * 7) +
                    (parseInt(number[17]) * 3) +
                    (parseInt(number[18]));
            alert(((sum2 % 10).toFixed(0)));
            var checkDigit2 = ((sum2 % 10).toFixed(0));

            if ((number[19]) === ("" + checkDigit2)) {
                alert('Our Checkdigit2 is valid', checkDigit2);
                valid = true;
            }
        } catch (e) {
            alert('Fail for check 2!');
            valid = false;
        }
        try {
            var sum3 = (parseInt(number[21]) * 7) +
                    (parseInt(number[22]) *3) +
                    (parseInt(number[23])) +
                    (parseInt(number[24]) * 7) +
                    (parseInt(number[25]) * 3) +
                    (parseInt(number[26]));
            alert(((sum3 % 10).toFixed(0)));
            var checkDigit3 = ((sum3 % 10).toFixed(0));

            if ((number[27]) === ("" + checkDigit3)) {
                alert('Our Checkdigit3 is valid',checkDigit3);
                valid = true;
            }
        } catch (e) {
            valid = false;
        }

        alert('All Good DUde!');
        return valid;
    }

Answer 1

From experience, you may want to separate as much math as possible in your try block. JavaScript has a weird way of handling variables and may not be doing what you think it is.

Answer 2

Here is the way to do it.

I have not thrown any error, only error can be if the number is not parseable and so you can throw it if you like else if your sum checks can validate that should be good enough

window.onkeyup = keyup;
var number;

function keyup(e) {
    number = e.target.value;
    $('#numberValue').text(number);
    // must be 10 characters long
    if (number.length !== 30) {
        return false;
    }
    number = "" + (number || "");
    var valid = false;
    //try{
        var sum1 = returnSum(number,[0,1,2,3,4,5,6,7,8],[7,3,1,7,3,1,7,3,1]);
        var sum2 = returnSum(number,[13,14,15,16,17,18],[7,3,1,7,3,1]);
        var sum3 = returnSum(number,[21,22,23,24,25,26],[7,3,1,7,3,1]);
    /*
    //only if you are throwing err
    }catch(e){
        valid = false;
    }
    */

    if (number[9] === sum1 && number[19] === sum2 && number[27] === sum3) {
        console.log(sum1 +'|' + sum2 + '|' + sum3);
        valid = true;
    }


    console.log('All Good DUde!');
    return valid;
}

function myParse(n){
    return (isNaN(parseInt(n,10))) ? -1 : parseInt(n,10);
}

function returnSum(n,ind,mul){
    var acc = 0;
    var pNum = 0;
    for(var i=0; i<ind.length; i++){
        pNum = myParse(n[ind[i]]);
        if(pNum == -1){
            pNum=0;
            //throw 'error';//if you really want to throw error on not a number / or your number should fail
        }
        acc += pNum * mul[i];
    }
    return (acc%10).toFixed(0)+'';
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<h3>  Sample test number to use -- copy and paste should work </p=h3>
<p>487013675311199070109160101300</p>

<input id="searchTxt" placeholder="add numbers together">
<div id='numberValue'>Number goes here</div>

Cheers. joy