CODEWITHSUNDEEP
javascriptjquery
user892134
user892134

Reputation: 3224

jquery adding decimal numbers

I'm trying to add together decimal numbers but when i alert the variable finalnumber i get zero. The variable number is a decimal number. How do i solve this so that variable finalnumber is the sum of all number?

var finalnumber = 0;

$('#chosen-keyword-container').find('.keyword-row').each(function() {
    var number = $(this).find('td:last').find('input[name=bid-price[]]').val();

    var finalnumber = parseInt(number) + parseInt(finalnumber);

});​

Upvotes: 3

Views: 16569

Answers (3)

gdoron
gdoron

Reputation: 150253

Change this:

var finalnumber = parseInt(number)+parseInt(finalnumber);

To this:

finalnumber = finalnumber + parseFloat(number);

Or:

finalnumber += parseFloat(number);
  • parseInt can't hold decimal values. use parseFloat instead.
  • Don't declare finalnumber with var, becuase it hides the finalnumber in the outer scope.

Upvotes: 10

rt2800
rt2800

Reputation: 3045

change the addition line as follows

finalnumber = finalnumber + parseFloat(number);

Upvotes: 1

Sirko
Sirko

Reputation: 74046

Just drop the var keyword inside your function in front of finalnumber. With that var you define a new variable under that name and scope. So basically you have two versions of finalnumber and just add to the local one (the one inside the function and not the global one).

On the other hand you should change parseInt to parseFloat as you are working with decimal numbers (See answer of @gdoron).

$('#chosen-keyword-container').find('.keyword-row').each(function() {
  var number = $(this).find('td:last').find('input[name=bid-price[]]').val();

  finalnumber = parseFloat(number) + finalnumber;  
});

On a sidenote: You can drop the parseInt() inside the function for finalnumber. This variable is always a number value and so there is no need to convert it.

Upvotes: 2

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