Question

This is my javascript code :

var dsum = 0.0;
$.each($("#Report tbody tr"), function () {
    var x = 0.0;

    if ($(this).find("td:eq(3) div").html() != " ")
        x = parseFloat($(this).find("td:eq(3) div").html().replace(/,/g, ""));

    dsum += x;

});

var csum = 0.0;

$.each($("#Report tbody tr"), function () {
     var x = 0.0;
     if ($(this).find("td:eq(4) div").html() != " ")
         x = parseFloat($(this).find("td:eq(4) div").html().replace(/,/g, ""));
         csum += x;

});


var  lastEvenRow=$("#Report").find("tr:last").hasClass('erow');
$("<tr id='LastRow'><td colspan='3' style='border-top: 1px solid #DDDDDD;'><div style='padding:5px;text-align:right;'>Total :</div></td><td style='text-align:center;height:22px;border-top: 1px solid #DDDDDD;'><div style='padding:5px;'>" + dsum.toString(10).replace( /(\d)(?=(\d\d\d)+(?!\d))/g , "$1,") + "</div></td><td style='text-align:center;border-top: 1px solid #DDDDDD;'><div style='padding:5px;'>" + csum.toString(10).replace( /(\d)(?=(\d\d\d)+(?!\d))/g , "$1,") + "</div></td><td style='text-align:center;border-top: 0px solid #DDDDDD;'><div style='padding:5px;border-top: 1px solid #DDDDDD;'>" + (dsum > csum ? (dsum - csum).toString(10).replace( /(\d)(?=(\d\d\d)+(?!\d))/g , "$1,") : "&nbsp;") + "</div></td><td style='text-align:center;'><div style='padding:5px;'>" + (csum >= dsum ? (csum - dsum).toString(10).replace( /(\d)(?=(\d\d\d)+(?!\d))/g , "$1,") : "&nbsp;") + "</div></td></tr>").appendTo("#Report tbody");
if(lastEvenRow==false)
    $("#LastRow").addClass("erow");

As I mentioned in the picture , i think the result must be something like 104,889.01 but , as you see , It is Incorrect value

Is there any body out there to help me ?
Sorry about my bad English syntax(I am new in English)

UPDATE1 :

1318114.01 – 1213225 must be 104889.01 , so why javascript return 104889.01000000001?!!

Answer 1

It happens because it works with convertions from binary to decimal, then we have aproximations, whose result on big floating numbers. See:

var i = 1;//1
i += .1;//1.1
i += .1;//1.2000000000000002

You can try an own aproximation:

var result = Math.round((1318114.01-1213225)*1000)/1000;