Javascript String to int conversion

Question

I have the following JS immbedded in a page:

var round = Math.round;
var id = $(this).attr("id");

var len = id.length;
var indexPos = len -1; // index of the number so that we can split this up and used it as a title

var pasType = id.substring(0, indexPos); // adult, child or infant
var ind = round(id.substring(indexPos)); // converts the string index to an integer
var number = (id.substring(indexPos) + 1); // creates the number that will go in the title
window.alert(number);

id will be something like adult0, and I need to take that string and split it into adult and 0 - this part works fine.

The problem comes in when I try to increment the 0. As you can see I use Math.round to convert it to an integer, and then add 1 to it - I expect 0 to be 1 after this. However, it doesn't seem to be converting it to integer, because I get 01, not 1. When testing this with adult1 the alert I get is 11.

I'm using this question for reference, and have also tried var number += id.substring(indexPos);, which breaks the JS (unexpected identifier '+=')

Does anyone know what I'm doing wrong? Is there a better way of doing this?

You have already converted a number - just use it: number = ind + 1. — georg
– georg, Commented Sep 10, 2013 at 8:07

Charaf JRA · Accepted Answer · 2014-02-28 20:49:22Z

76

The parseInt() function parses a string and returns an integer,10 is the Radix or Base [DOC]

var number = parseInt(id.substring(indexPos) , 10 ) + 1;

edited Feb 28, 2014 at 20:49

answered Sep 10, 2013 at 8:03

Charaf JRA

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1 Comment

Factor Mystic Over a year ago

FYI: always pass the radix parameter to parseInt developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Arshid KV · Accepted Answer · 2018-01-29 06:52:44Z

17

Convert by Number Class:-

Eg:

var n = Number("103");
console.log(n+1)

Output: 104

Note:- Number is class. When we pass string, then constructor of Number class will convert it.

answered Jan 29, 2018 at 6:52

Arshid KV

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2 Comments

mKane Over a year ago

But if the string has extra chars in it then it will be NaN.

Abdulaziz Hamdan Over a year ago

Working with a string of floating numbers, this worked for me. Simple and straightforward!

Qantas 94 Heavy · Accepted Answer · 2013-12-04 09:37:08Z

This is to do with JavaScript's + in operator - if a number and a string are "added" up, the number is converted into a string:

0 + 1; //1
'0' + 1; // '01'

To solve this, use the + unary operator, or use parseInt():

+'0' + 1; // 1
parseInt('0', 10) + 1; // 1

The unary + operator converts it into a number (however if it's a decimal it will retain the decimal places), and parseInt() is self-explanatory (converts into number, ignoring decimal places).

The second argument is necessary for parseInt() to use the correct base when leading 0s are placed:

parseInt('010'); // 8 in older browsers, 10 in newer browsers
parseInt('010', 10); // always 10 no matter what

There's also parseFloat() if you need to convert decimals in strings to their numeric value - + can do that too but it behaves slightly differently: that's another story though.

Jite · Accepted Answer · 2013-09-10 08:17:38Z

10

JS will think that the 0 is a string, which it actually is, to convert it to a int, use the: parseInt() function, like:

var numberAsInt = parseInt(number, 10);  
// Second arg is radix, 10 is decimal.

If the number is not possible to convert to a int, it will return NaN, so I would recommend a check for that too in code used in production or at least if you are not 100% sure of the input.

edited Sep 10, 2013 at 8:17

answered Sep 10, 2013 at 8:04

Jite

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Comments

Alexander Jank · Accepted Answer · 2017-03-08 07:59:47Z

6

Although parseInt is the official function to do this, you can achieve the same with this code:

number*1

The advantage is that you save some characters, which might save bandwidth if your code has to lots of such conversations.

answered Mar 8, 2017 at 7:59

Alexander Jank

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1 Comment

Dan Over a year ago

Note that the behaviour of number*1 is not the same as parseInt for all inputs.

Charaf JRA · Accepted Answer · 2013-09-10 08:08:33Z

3

Use parseInt():

var number = (parseInt(id.substring(indexPos)) + 1);` // creates the number that will go in the title

edited Sep 10, 2013 at 8:08

Charaf JRA

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answered Sep 10, 2013 at 8:05

juju

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user1203684 · Accepted Answer · 2013-09-10 08:04:22Z

2

If you are sure id.substring(indexPos) is a number, you can do it like so:

var number = Number(id.substring(indexPos)) + 1;

Otherwise I suggest checking if the Number function evaluates correctly.

answered Sep 10, 2013 at 8:04

user1203684

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