var a = 1;
var b = 2;
var c = a+b;
c will show as 12; but I need 3
How do I do it using jQuery?
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4Not sure if serious :/Gogol– Gogol2020-07-26 15:06:31 +00:00Commented Jul 26, 2020 at 15:06
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6 Answers
Definitely use jQuery
Because the power of jQuery is obviously unparalleled, here is how to do with with 100% jQuery:
var a = 1;
var b = 2;
$("<script>c=+("+a+")+ +("+b+")</script>").appendTo($(document));
Now c will hold your result and you used nothing but jQuery! As you can see jQuery is really great because it does all sorts of things
This also works well because it doesn't matter if a or b are strings!
Not enough jQuery?
var a = 1;
var b = 2;
$("<script id='test'>$('<textarea id=\\'abc\\'>'+("+a+")+ +("+b+")+'</textarea>').appendTo($('body'))</script>").appendTo('body');
var c = $("#abc").val();
This answer was done with 100% jQuery because jQuery is awesome but only use this carefully because it might not work sometimes.
jQuery Arithmetic Plugin
You can also use the revolutionary jQuery Arithmetic Plugin this has solved world peace in over 4294967295 (>> 0 === -1) countries:
var a = 1;
var b = 2;
var c = $.add(a,b);
while this is all great (this is not confirmed), but in jQuery -3.0.1 I have heard you will be able to add numbers this way:
$.number($.one, $.two).add($.number($.three, $.four))
this adds 12 (one + two) to 34 (three + four)
1 Comment
ARI FISHER
This is awesome and hilarious.
It looks like you have strings and not numbers, you need parseInt() or parseFloat() (if they may be decimals) here, like this:
var a = "1";
var b = "2";
var c = parseInt(a, 10) + parseInt(b, 10);
//or: var c = parseFloat(a) + parseFloat(b);
You can test the difference here, it's worth noting these are not jQuery but base JavaScript functions, so this isn't dependent on the jQuery library in any way.
3 Comments
Ashnur
I had to downvote this, because you should never use parseInt to cast strings to number. Rather, use Number().
Armen Michaeli
Maybe, just maybe, because of implications like discussed at stackoverflow.com/questions/4090518/…
Roy Tinker
Not enough jQuery.
Just try this:
var a = 1;
var b = 2;
var c = (+a) + (+b);
alert(c); //or whatever you want
Comments
Try this -
var c = parseInt(a, 10) + parseInt(b, 10);
2 Comments
Nick Craver
You should always use a radix on
parseInt(), otherwise numbers starting with a 0 will be treated as base 16 by default.
Douglas
Watch out with this one, it will use "0" as a magic prefix for octal, so:
parseInt("010") == 8.This performs the addition of two variables supplied from input variables selected on the basis of ID.
var salary, tds, netSalary;
salary = parseInt($("#txtSalary").val());
$("#txtTds").on('mouseenter focus', function ()
{
tds = parseInt($("#txtSalary").val() * (0.1));
$("#txtTds").val(tds);
});
$("#txtNetSalary").on('mouseenter focus', function () {
netSalary = parseInt($("#txtSalary").val() +("#txtTds").val());
$("#txtNetSalary").val(netSalary);
});
Comments
var a =10;
var b=10;
var c=a+b;
alert(c);
//in double case
var e=10.5;
var f=10.5;
alert(e+f);
//forcelly convert to int
alert(parseInt(e)+parseInt(f));
//forcelly convert to float
alert(parseFloat(a)+parseFloat(b));
Example :
https://jsfiddle.net/v0rjek67/6/
