JavaScript parseFloat() issue

Question

Okay so my code works fine but when a decimal i.e. 60.1, 60.2, 60.3, etc. is input for #chance it screws up profit and pay.

For example: 60% is input for chance, 1 for bet. It returns 1.65 for pay and 0.65 for profit. That's all correct.

But when I put 60.1, it returns 16.5 ( wrong decimal ) and 15.5 for profit. 16.5 seems like an easy fix but Idk how to fix it, but I have no idea why it's returning 15.5 for profit and thought maybe if I fixed pay it would fix the issue with profit.

What's wrong?

Thanks.

<script>
    $(document).ready(function(){

        function updateValues() {
            // Grab all the value just incase they're needed.
            var chance = $('#chance').val();
            var bet = $('#bet').val();
            var pay = $('#pay').val();
            var profit = $('#profit').val();

            // Calculate the new payout.
            var remainder = 101 - chance;
            pay = Math.floor((992/parseFloat((chance+0.5))) *100)/100;


            // Calculate the new profit.
            profit = bet*pay-bet;
                            profit = profit.toFixed(6);


            // Set the new input values.
            $('#chance').val(chance);
            $('#bet').val(bet);
            $('#pay').val(pay);
            $('#profit').val(profit);
        }


        parseInt($('#chance').keyup(updateValues));
        parseInt($('#bet').keyup(updateValues));
        parseInt($('#pay').keyup(updateValues));
        parseInt($('#profit').keyup(updateValues));


    });
</script>

WTH is parseInt($('#…').keyup(updateValues)) supposed to do? You're parsing jQuery wrapper objects but don't use the NaN results then. — Bergi
– Bergi, Commented Jun 27, 2013 at 0:53
Also, "set the new input values" - only two of them changed; two of the lines are useless. — Amadan
– Amadan, Commented Jun 27, 2013 at 0:54

Bergi · Accepted Answer · 2013-06-27 01:00:14Z

2

parseFloat((chance+0.5))

looks very wrong. chance is a string, so the + operator will perform string concatenation. When you input 60, it becomes

parseFloat("60"+0.5) === 600.5

while when you input 60.1 it is

parseFloat("60.1"+0.5) === 60.1

You probably wanted

(992/(parseFloat(chance)+0.5))*100
// or
(992/parseFloat(chance)+0.5)*100
// or
992/parseFloat(chance)*100+0.5
// or something along these lines

edited Jun 27, 2013 at 1:00

answered Jun 27, 2013 at 0:51

Bergi

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5 Comments

Jake McLaughlin Over a year ago

Wrong: puu.sh/3paY5.png Right: puu.sh/3paYU.png Is that helping you understand?

Bergi Over a year ago

@Jake: Not really. Please tell me about the maths you are attempting to use.

Jake McLaughlin Over a year ago

Ok so I'm making a gambling site and it's a 2% house edge. So the pay should return 2 flat if 49 is entered for chance. 2% (51-49). And when they enter 60, it should give the proper X(multiplier) for what they should be paid out factoring in the house edge. Except when they enter a decimal it screws everything up.

Jake McLaughlin Over a year ago

pay = Math.floor((9900/(parseFloat(chance)+0.5))*100)/10000; This works ^. Thanks everyone. BUT, how can I make it so they can change pay and it will update chance? Thanks.

Bergi Over a year ago

@Jake: Yeah, your maths was totally screwed up :-) To have different changing fields, you will need a different handler that does the math in reverse

Amadan · Accepted Answer · 2013-06-27 00:49:23Z

1

Change parseFloat((chance+0.5)) into (parseFloat(chance)+0.5).

Actually, I can't see why it's working with 60, either. chance, as a value of a text field, is a string: "60". Strings don't add, they concatenate: "60" + 0.5 is "600.5", same as "60" + "0.5".

answered Jun 27, 2013 at 0:49

Amadan

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4 Comments

Jake McLaughlin Over a year ago

Didn't fix the problem. I did that though.

Amadan Over a year ago

Well, that's the right thing to do: you got two independent answers saying the very same thing. If it still doesn't work, update the question and show what exactly is happening.

Jake McLaughlin Over a year ago

This is me updating it telling you I still have the same problem as posted in the OP.

Amadan Over a year ago

And I'm telling you that your "update" is useless, because there is no way in hell that you can get the same result if you changed what @Bergi and I told you. Run this and look at the console - for all three test values, the result is changed. So while you might still have a problem, either you have a different problem, or you have applied the change wrong, but in either case you need to update your code, not say "doesn't work".

StackSlave · Accepted Answer · 2013-06-27 01:54:53Z

Try something like this:

$(document).ready(function(){
  function updateValues(){
    var chance = $('#chance').val();
    var bet = $('#bet').val();
    var pay = $('#pay').val();
    var profit = $('#profit').val();
    pay = ((992/Math.floor(+chance+0.5))/10).toFixed(2);
    profit = (bet*pay-bet).toFixed(6);
    $('#chance').val(chance);
    $('#bet').val(bet);
    $('#pay').val(pay);
    $('#profit').val(profit);
  }
  $('#chance').keyup(updateValues);
  $('#bet').keyup(updateValues);
  $('#pay').keyup(updateValues);
  $('#profit').keyup(updateValues);
});

Something is wrong with your Math.

Note:

You don't have to use parseInt() or parseFloat() to make Strings to Numbers. the + symbol in front of your String that is a Number will convert it to a Number.

See http://jsfiddle.net/PHPglue/JQJMD/ for more details.

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