JavaScript: Get the second digit from a number? - Stack Overflow

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I have a number assigned to a variable, like that:

var myVar = 1234;

Now I want to get the second digit (2 in this case) from that number without converting it to a string first. Is that possible?

I have a number assigned to a variable, like that:

var myVar = 1234;

Now I want to get the second digit (2 in this case) from that number without converting it to a string first. Is that possible?

• javascript
• numbers

Share Improve this question Follow edited Dec 19, 2012 at 15:33 John Kugelman 361k6969 gold badges546546 silver badges592592 bronze badges asked Dec 19, 2012 at 15:31 user1856596user1856596 7,2231313 gold badges4444 silver badges6565 bronze badges 3

• //var secondNumber = (myVar-1000)/100; – BLOB Commented Dec 19, 2012 at 15:34
• 5 @BLOB I have never seen a digit with a fraction part ;) – Rob W Commented Dec 19, 2012 at 15:34
• 3 Why without converting it to a string? – Florian Margaine Commented Dec 19, 2012 at 15:34

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14 Answers 14

Sorted by: Reset to default 94

So you want to get the second digit from the decimal writing of a number.

The simplest and most logical solution is to convert it to a string :

var digit = (''+myVar)[1];

or

var digit = myVar.toString()[1];

If you don't want to do it the easy way, or if you want a more efficient solution, you can do that :

var l = Math.pow(10, Math.floor(Math.log(myVar)/Math.log(10))-1);
var b = Math.floor(myVar/l);
var digit = b-Math.floor(b/10)*10;

Demonstration

For people interested in performances, I made a jsperf. For random numbers using the log as I do is by far the fastest solution.

1st digit of number from right → number % 10 = Math.floor((number / 1) % 10)

1234 % 10; // 4
Math.floor((1234 / 1) % 10); // 4

2nd digit of number from right → Math.floor((number / 10) % 10)

Math.floor((1234 / 10) % 10); // 3

3rd digit of number from right → Math.floor((number / 100) % 10)

Math.floor((1234 / 100) % 10); // 2

nth digit of number from right → Math.floor((number / 10^n-1) % 10)

function getDigit(number, n) {
  return Math.floor((number / Math.pow(10, n - 1)) % 10);
}

number of digits in a number → Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1 Credit to: https://stackoverflow.com/a/28203456/6917157

function getDigitCount(number) {
  return Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1;
}

nth digit of number from left or right

function getDigit(number, n, fromLeft) {
  const location = fromLeft ? getDigitCount(number) + 1 - n : n;
  return Math.floor((number / Math.pow(10, location - 1)) % 10);
}

You know, I get that the question asks for how to do it without a number, but the title "JavaScript: Get the second digit from a number?" means a lot of people will find this answer when looking for a way to get a specific digit, period.

I'm not bashing the original question asker, I'm sure he/she had their reasons, but from a search practicality standpoint I think it's worth adding an answer here that does convert the number to a string and back because, if nothing else, it's a much more terse and easy to understand way of going about it.

let digit = Number((n).toString().split('').slice(1,1))

// e.g.
let digit = Number((1234).toString().split('').slice(1,1)) // outputs 2

Getting the digit without the string conversion is great, but when you're trying to write clear and concise code that other people and future you can look at really quick and fully understand, I think a quick string conversion one liner is a better way of doing it.

Get rid of the trailing digits by dividing the number with 10 till the number is less than 100, in a loop. Then perform a modulo with 10 to get the second digit.

if (x > 9) {
    while (x > 99) {
        x = (x / 10) | 0;  // Use bitwise '|' operator to force integer result.
    }
    secondDigit = x % 10;
}
else {
    // Handle the cases where x has only one digit.
}

A "number" is one thing.

The representation of that number (e.g. the base-10 string "1234") is another thing.

If you want a particular digit in a decimal string ... then your best bet is to get it from a string :)

Q: You're aware that there are pitfalls with integer arithmetic in Javascript, correct?

Q: Why is it so important to not use a string? Is this a homework assignment? An interview question?

function getNthDigit(val, n){
    //Remove all digits larger than nth
    var modVal = val % Math.pow(10,n);

    //Remove all digits less than nth
    return Math.floor(modVal / Math.pow(10,n-1));
}

// tests
[
  0, 
  1, 
  123, 
  123456789, 
  0.1, 
  0.001
].map(v => 
  console.log([
      getNthDigit(v, 1),
      getNthDigit(v, 2),
      getNthDigit(v, 3)
    ]
  ) 
);

This is how I would do with recursion

function getDigits(n, arr=[]) {
   arr.push(n % 10)

   if (n < 10) {
     return arr.reverse()
   }
   return getDigits(Math.floor(n/10),arr)
  }

const arr = getDigits(myVar)
console.log(arr[2])

Just a simple idea to get back any charter from a number as a string or int:

const myVar = 1234;
String(myVar).charAt(1)
//"2"
parseInt(String(myVar).charAt(1))
//2

I don’t know why you need this logic, but following logic will get you the second number

<script type="text/javascript">
    var myVal = 58445456;
    var var1 = new Number(myVal.toPrecision(1));
    var var2 = new Number(myVal.toPrecision(2));     
    var rem;
    rem = var1 - var2;
    var multi = 0.1;
    var oldvalue;
    while (rem > 10) {
        oldvalue = rem;
        rem = rem * multi;
        rem = rem.toFixed();           
    }
    alert(10-rem);       
</script>

function getDigit(number, indexFromRight) { 
            var maxNumber = 9
            for (var i = 0; i < indexFromRight - 2; i++) {
                maxNumber = maxNumber * 10 + 9
            }
            if (number > maxNumber) {
                number = number / Math.pow(10, indexFromRight - 1) | 0
                return number % 10
            } else
                return 0
        }

you can use this function index = 0 will give you the first digit from the right (the ones) index = 1 will give you the second digit from the right (the tens)

and so on

const getDigit = (num, index) => {
    

    if(index === 0) {
        return num % 10;
    }

    let result = undefined;

    for(let i = 1; i <= index; i++) {
        num -= num % 10;
        num /= 10;
        result = num % 10;
    }

    return result;

}

for Example:

getDigit(125, 0)   // returns 5
gitDigit(125, 1)   // returns 2
gitDigit(125, 2)   // returns 1
gitDigit(125, 3)   // returns 0

function left(num) {
    let newarr = [];
    let numstring = num.split('[a-z]').join();
    //return numstring;
    const regex = /[0-9]/g;
    const found = numstring.match(regex);
   // return found;
    for(i=0; i<found.length; i++){
        return found[i];
    }
    } 
//}
console.log(left("TrAdE2W1n95!"))

function getNthDigit(n, number){
    return ((number % Math.pow(10,n)) - (number % Math.pow(10,n-1))) / Math.pow(10,n-1);
}

Explanation (Number: 987654321, n: 5):
a = (number % Math.pow(10,n)) - Remove digits above => 54321
b = (number % Math.pow(10,n-1)) - Extract digits below => 4321
a - b => 50000
(a - b) / 10^(5-1) = (a - b) / 10000 => 5

var newVar = myVar;
while (newVar > 100) {
    newVar /= 10;
}

if (newVar > 0 && newVar < 10) {
   newVar = newVar;
}

else if (newVar >= 10 && newVar < 20) {
   newVar -= 10;
}

else if (newVar >= 20 && newVar < 30) {
   newVar -= 20;
}

else if (newVar >= 30 && newVar < 40) {
   newVar -= 30;
}

else if (newVar >= 40 && newVar < 50) {
   newVar -= 40;
}

else if (newVar >= 50 && newVar < 60) {
   newVar -= 50;
}

else if (newVar >= 60 && newVar < 70) {
   newVar -= 60;
}

else if (newVar >= 70 && newVar < 80) {
   newVar -= 70;
}

else if (newVar >= 80 && newVar < 90) {
   newVar -= 80;
}

else if (newVar >= 90 && newVar < 100) {
   newVar -= 90;
}

else {
   newVar = 0;
}

var secondDigit = Math.floor(newVar);

That's how I'd do it :)

And here's a JSFiddle showing it works :) http://jsfiddle.net/Cuytd/

This is also assuming that your original number is always greater than 9... If it's not always greater than 9 then I guess you wouldn't be asking this question ;)

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