Develop Reference

Javascript convert string to integer

I am just dipping my toe into the confusing world of javascript, more out of necessity than desire and I have come across a problem of adding two integers.
1,700.00 + 500.00
returns 1,700.00500.00
So after some research I see that 1,700.00 is being treated as a string and that I need to convert it.
The most relevant pages I read to resolve this were this question and this page. However when I use
parseInt(string, radix)
it returns 1. Am I using the wrong function or the an incorrect radix (being honest I can't get my head around how I decide which radix to use).
var a="1,700.00";
var b=500.00;
parseInt(a, 10);

Basic Answer
The reason parseInt is not working is because of the comma. You could remove the comma using a regex such as:
var num = '1,700.00';
num = num.replace(/\,/g,'');
This will return a string with a number in it. Now you can parseInt. If you do not choose a radix it will default to 10 which was the correct value to use here.
num = parseInt(num);
Do this for each of your string numbers before adding them and everything should work.
More information
How the replace works:
More information on replace at mdn:
`/` - start
`\,` - escaped comma
`/` - end
`g` - search globally
The global search will look for all matches (it would stop after the first match without this)
'' replace the matched sections with an empty string, essentially deleting them.
Regular Expressions
A great tool to test regular expressions: Rubular and more info about them at mdn
If you are looking for a good tutorial here is one.
ParseInt and Rounding, parseFloat
parseInt always rounds to the nearest integer. If you need decimal places there are a couple of tricks you can use. Here is my favorite:
2 places: `num = parseInt(num * 100) / 100;`
3 places: `num = parseInt(num * 1000) / 1000;`
For more information on parseInt look at mdn.
parseFloat could also be used if you do not want rounding. I assumed you did as the title was convert to an integer. A good example of this was written by #fr0zenFry below. He pointed out that parseFloat also does not take a radix so it is always in base10. For more info see mdn.

Try using replace() to replace a , with nothing and then parseFloat() to get the number as float. From the variables in OP, it appears that there may be fractional numbers too, so, parseInt() may not work well in such cases(digits after decimal will be stripped off).
Use regex inside replace() to get rid of each appearance of ,.
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = a+b;
This should give you correct result even if your number is fractional like 1,700.55.
If I go by the title of your question, you need an integer. For this you can use parseInt(string, radix). It works without a radix but it is always a good idea to specify this because you never know how browsers may behave(for example, see comment #Royi Namir). This function will round off the string to nearest integer value.
var a = parseInt('1,700.00'.replace(/,/g, ''), 10); //radix 10 will return base10 value
var b = parseInt('500.00'.replace(/,/g, ''), 10);
var sum = a+b;
Note that a radix is not required in parseFloat(), it will always return a decimal/base10 value. Also, it will it will strip off any extra zeroes at the end after decimal point(ex: 17500.50 becomes 17500.5 and 17500.00 becomes 17500). If you need to get 2 decimal places always, append another function toFixed(decimal places).
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = (a+b).toFixed(2); //change argument in toFixed() as you need
// 2200.00
Another alternative to this was given by #EpiphanyMachine which will need you to multiply and then later divide every value by 100. This may become a problem if you want to change decimal places in future, you will have to change multiplication/division factor for every variable. With toFixed(), you just change the argument. But remember that toFixed() changes the number back to string unlike #EpiphanyMachine solution. So you will be your own judge.

try this :
parseFloat(a.replace(/,/g, ''));
it will work also on : 1,800,300.33
Example :
parseFloat('1,700,800.010'.replace(/,/g, '')) //1700800.01

Javascript doesn't understand that comma. Remove it like this:
a.replace(',', '')
Once you've gotten rid of the comma, the string should be parsed with no problem.

New to Javascript calculations

I'm BRAND new to javascript so this is probably an easy fix I can't figure out.
I'm writing a easy script to calculate age Leapyear, Dog years and plus Five.
var age = prompt ("what's your age?");
var dog = age + 7;
var leapyear = age / 4;
var plusFive = age + 5;
document.write(dog, leapyear, plusFive);
JS isn't calculating age + 7. It's writing the age + 7. So if I write 15 in the age prompt it will print in the browser 157.(15) and (7) Which i understand and know why. but how do I get it to compute the math.
It's actually returning 1573.75155
thanks

Like what everyone's been saying, prompt returns a string so it needs to be converted. There are a number of ways to do this, some of which have already been mentioned:
parseInt('123')
parseFloat('123')
Number('123')
These are probably the most common and depending on context also quite possibly the most clear and intuitive ways of doing it. There are also a couple of very terse and interesting ways of converting strings to numbers, depending on which kind of number you'd like. For instance, to convert a number in a string to a float, you can prefix it with the + operator:
+'1.23'
This can seem really counter intuitive, particularly since 4 + '1.23' will actually return 41.23. So what's going on? Well, the + operator, when used as a unary operator (that is, it only has one operand on the right hand side) will always convert the operand value to a number. Compare these two (try them in a javascript console):
4 + '1.23' // returns 41.23
4 + +'1.23' // returns 5.23; the second + is a unary operator
In contrived examples such as this, it really reads rather badly so you might not want to use this trick everywhere. But in your code, it reads quite well:
var age = +prompt("What's your age?")
var dog = age + 7;
var leapyear = age / 4;
var plusFive = age + 5;
If you understand the workings of the unary plus operator (it really isn't rocket surgery) then you can get some nice terse but quite comprehensible results.
The unary + operator will always convert the value to a Number, i.e. floating point. Now, you might want an integer instead, in which case you can use the bitwise not operator twice, like so:
4 + ~~'1.23' // returns 5; note the double operator
This operator first converts the value to an integer, and then returns the bitwise complement of the value, meaning all the bits are inverted. Using it twice will mean that we get the complement's complement, i.e. the original value but this time as an integer instead of a float.
Hope this was informative!

Right now Javascript handles the input as a string, so age is a string. You're gonna want to convert that to an int using the function parseInt(age).
Official documentation
Also, I'd suggest you read this about types in JS

Use either parseInt(age) or parseFloat(age) depending on whether you want to accept non-integer ages.

Your prompt is returning your number as a string, so when you calculate 15 + 7 it's actually just concatenating "7" on to "15".
You need to convert your string to an number:
var age = prompt ("what's your age?");
age = parseInt(age,10); // Converts to an integer
// parseFloat(age) will allow fractional ages
var dog = age + 7;
var leapyear = age / 4;
var plusFive = age + 5;
document.write(dog, leapyear, plusFive);

the age is going to be treated as string so string plus int result in string you have to convert age to int :
var age= parseInt(prompt("what's your age"));
updated...

Javascript is treating variables as Strings, why?

I have the variable y, which is a subtotal. Its value is different depending on what happens with the html, but throughout the script I declared it like this:
var y = 21.78;
etc. Why is it that on my last equation where I add up the total, it treats them as strings when I want to add the values?
var tax = (0.055*y).toFixed(2);
var totalprice = y+tax;
/* totalprice holds "21.781.20" instead of 22.98 */

According to:
https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Number/toFixed
toFixed() returns:
A string representation of number that does not use exponential
notation and has exactly digits digits after the decimal place.
thus, y+tax is cast to a string since one of the operands is a string.
In my opinion, this would make sense as Javascript's intrinsic numeric types do not have the ability to store a specific number of decimal place digits, so a string would be the most appropriate data structure to store this with.
I would advise you do all your addition before calling toFixed(), since the method is most suitable for formatting display output.
var taxRate = 0.055;
var subtotal = 21.78;
var tax = (taxRate * subtotal).toFixed(2),
totalprice = ((1+taxRate) * subtotal).toFixed(2);
document.write(totalprice);

The .toFixed() method returns a string. Try applying that method as the last step after all other calculations.

Here's a simple fix. Put '+' in front of the tax variable to convert it to a number.
var y = 21.78;
var tax = (0.055*y).toFixed(2);
var totalprice = y+ (+tax);
totalprice === 22.98;
If you don't want any rounding errors when you use toFixed, then include this re-implementation of it in your script.
http://bateru.com/news/2012/03/reimplementation-of-number-prototype-tofixed/

In my experience, if there's any chance available, Javascript will see the "+" sign as concatenate rather than addition. It's driven me nuts on more than one occasion. I will generally do this rather than chance concatenation:
var totalprice = parseInt(y)+parseInt(tax);

When letter replaces value, multiply with 1 when you're in need of +.
var totalprice = (y*1) + tax .
Other operands work fine, it's just the + operand that needs special treatment when variable replace value.

Preventing concatenation

I've been writing JavaScript on and off for 13 years, but I sort of rediscovered it in the past few months as a way of writing programs that can be used by anyone visiting a web page without installing anything. See for example.
The showstopper I've recently discovered is that because JavaScript is loosely typed by design, it keeps concatenating strings when I want it to add numbers. And it's unpredictable. One routine worked fine for several days then when I fed different data into it the problem hit and I ended up with an impossibly big number.
Sometimes I've had luck preventing this by putting ( ) around one term, sometimes I've had to resort to parseInt() or parseFloat() on one term. It reminds me a little of trying to force a float result in C by putting a .00 on one (constant) term. I just had it happen when trying to += something from an array that I was already loading by doing parseFloat() on everything.
Does this only happen in addition? If I use parseInt() or parseFloat() on at least one of the terms each time I add, will that prevent it? I'm using Firefox 6 under Linux to write with, but portability across browsers is also a concern.

The specification says about the addition operator:
If Type(lprim) is String or Type(rprim) is String, then
Return the String that is the result of concatenating ToString(lprim) followed by ToString(rprim)
Which means that if at least one operator is a string, string concatenation will take place.
If I use parseInt() or parseFloat() on at least one of the terms each time I add, will that prevent it?
No, all operands have to be numbers.
You can easily convert any numerical string into a number using the unary plus operator (it won't change the value if you are already dealing with a number):
var c = +a + +b;

I normally do this:
var x = 2;
var t = "12";
var q = t+x; // q == "122"
var w = t*1+x; // *1 forces conversion to number w == 14
If t isn't a number then you'll get NaN.
If you multiply by 1 variables you don't know what type they are. They will be converted to a number. I find this method better than doing int and float casts, because *1 works with every kind of numbers.
The problem you are having is that the functions which fetch values from the DOM normally return strings. And even if it is a number it will be represented as a string when you fetch it.

You can use + operator to convert a string to number.
var x = '111'
+x === 111

Rest assured it is very predictable, you just need to be familiar with the operators and the data types of your input.
In short, evaluation is left-to-right, and concatenation will occur whenever in the presence of a string, no matter what side of the operation.
So for example:
9 + 9 // 18
9 + '9' // '99'
'9' + 9 // '99'
+ '9' + 9 // 18 - unary plus
- '9' + 9 // 0 - unary minus
Some ternary expressions:
9 + '9' + 9 // '999'
9 + 9 + '9' // '189'

why do I get 24 when adding 2 + 4 in javascript

I am trying this:
function add_things() {
var first = '2';
var second = '4';
alert(first + second);
}
But it gives me 24 instead of 6, what am I doing wrong?

You're concatenating two strings with the + operator. Try either:
function add_things() {
var first = 2;
var second = 4;
alert(first + second);
}
or
function add_things() {
var first = '2';
var second = '4';
alert(parseInt(first, 10) + parseInt(second, 10));
}
or
function add_things() {
var first = '2';
var second = '4';
alert(Number(first) + Number(second));
}
Note: the second is only really appropriate if you're getting strings from say a property or user input. If they're constants you're defining and you want to add them then define them as integers (as in the first example).
Also, as pointed out, octal is evil. parseInt('010') will actually come out as the number 8 (10 in octal is 8), hence specifying the radix of 10 is a good idea.

Try this:
function add_things() {
var first = 2;
var second = 4;
alert(first + second);
}
Note that I've removed the single quotes; first and second are now integers. In your original, they are strings (text).

That is one of the "Bad Parts" of JavaScript, as a loosely typed language, the addition and concatenation operator is overloaded.
JavaScript is loosely typed, but that doesn't mean that it has no data types just because a value of a variable, object properties, functions or parameters don't need to have a particular type of value assigned to it.
Basically there are three primitive data types:
boolean
number
string
null and undefined are two special cases, everything else are just variations of the object type.
JavaScript type-converts values of types into a type suitable for the context of their use (type coercion).
In your example were trying to add two objects of type string, so a concatenation occur.
You can "cast" or type convert the variables to number in many ways to avoid this problem:
var a = "2";
var b = "4";
// a and b are strings!
var sum = Number(a) + Number(b); // Number constructor.
sum = +a + +b; // Unary plus.
sum = parseInt(a, 10) + parseInt(b, 10); // parseInt.
sum = parseFloat(a) + parseFloat(b); // parseFloat.
This is I think a very common mistake, for example when reading user input from form elements, the value property of form controls is string, even if the character sequence that it contain represents a number (as in your example).
The "Bad Part" which I talk, is about the dual functionality of the + operator, overloaded to be used for both, numeric addition and string concatenation.
The operation that the + operator will do is determined completely by the context. Only if the both operands are numbers, the + operator perform addition, otherwise it will convert all of its operands to string and do concatenation.

The single quotes cause the values to be treated as characters instead of numbers. '2' + '4' = '24' in the same way that 'snarf' + 'blam' = 'snarfblam'.

You could also force the interpreter to perform arithmetic when dealing with numbers in string forms by multiplying the string by 1 (since multiplication can't be done on a string, it'll convert to a number if it can):
// fun with Javascript...
alert(first * 1 + second * 1);
But it's probably best to go with CMS's suggestion of using Number() to force the conversion, since someone will probably come along later and optimize the expression by removing the 'apparently unnecessary' multiply-by-one operations.