How to float number while fetching data as json in jquery?

Question

My question is about to float number in jquery. I looked other questions about it in this site but I couldn't find which I look.

In part of my code, I used $.post in jquery. I send request to database and fetch data as json format. In my database, some numbers' format is like 345,54565000000. So, I want to float numbers 5 digit after comma.

Part of my code is :

$.post("http://localhost/ajax.php",
      function(data) {
          var listdata;
          $.each(data, function(i,item){

              listdata += "<td>"+item.number+"</td>";

          });

            $(".result).empty().html(listdata);

    },"json"
);

Some of my trials are like:(which is not worked)

1.)

var number = (item.number).toFixed(5);
listdata += "<td>"+number+"</td>";

2.)

var number=item.number;
var new_number = number.toFixed(5);
listdata += "<td>"+new_number+"</td>";

Thank you for your replies.

number = number.toString().split("."); number[1].=substr(0, 5); number = (number[0] + "" + number[1]); — PitaJ
– PitaJ, Commented Dec 8, 2012 at 23:51
are you using a comma or a period? Rather, do you want the output as a comma or as a decimal? — Samuel Cook
– Samuel Cook, Commented Dec 8, 2012 at 23:56

adeneo · Accepted Answer · 2012-12-08 23:56:05Z

3

var number = 345.54565000000;

var parsedNumber = parseFloat(parseInt(number*100000,10)/100000);

FIDDLE;

If that seperator is in fact a comma, you'll have to replace it with .replace(',', '.')

answered Dec 8, 2012 at 23:56

adeneo

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Comments

Jules · Accepted Answer · 2012-12-08 23:48:59Z

I use this function:

function roundNumber(number,decimals) {
    var newString;// The new rounded number
    decimals = Number(decimals);
    if (decimals < 1) {
        newString = (Math.round(number)).toString();
    } else {
        var numString = number.toString();
        if (numString.lastIndexOf(".") == -1) {// If there is no decimal point
            numString += ".";// give it one at the end
        }
        var cutoff = numString.lastIndexOf(".") + decimals;// The point at which to truncate the number
        var d1 = Number(numString.substring(cutoff,cutoff+1));// The value of the last decimal place that we'll end up with
        var d2 = Number(numString.substring(cutoff+1,cutoff+2));// The next decimal, after the last one we want
        if (d2 >= 5) {// Do we need to round up at all? If not, the string will just be truncated
            if (d1 == 9 && cutoff > 0) {// If the last digit is 9, find a new cutoff point
                while (cutoff > 0 && (d1 == 9 || isNaN(d1))) {
                    if (d1 != ".") {
                        cutoff -= 1;
                        d1 = Number(numString.substring(cutoff,cutoff+1));
                    } else {
                        cutoff -= 1;
                    }
                }
            }
            d1 += 1;
        } 
        newString = numString.substring(0,cutoff) + d1.toString();
    }
    if (newString.lastIndexOf(".") == -1) {// Do this again, to the new string
        newString += ".";
    }
    var decs = (newString.substring(newString.lastIndexOf(".")+1)).length;
    for(var i=0;i<decimals-decs;i++) newString += "0";
    return newString;
}

So in your code, change line 6 to:

listdata += "<td>"+ roundNumber(item.number, 5) +"</td>";

Thank you for your reply. It works; but without this huge function, other replies does work. Thank you for your interest.

Marcos Besteiro López · Accepted Answer · 2012-12-08 23:56:52Z

1

I assume you have commas, and you want to stick with commas, so that will break any Float related functions, so... quick and dirty, but does the job:

Replace

listdata += "<td>"+item.number+"</td>";

With

listdata += "<td>"+parseFloat(item.number.replace(',', '.')).toFixed(5).replace('.', ',')+"</td>";

answered Dec 8, 2012 at 23:56

Marcos Besteiro López

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Comments

hobberwickey · Accepted Answer · 2012-12-09 15:41:51Z

1

the fasted way I've found is

((number * Math.pow(10, numberOfDigits)) | 0) / Math.pow(10, numberOfDigits)

EDIT: forgot the Math.pow, that's important.

edited Dec 9, 2012 at 15:41

answered Dec 9, 2012 at 0:05

hobberwickey

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