Ensuring `random` 3 digit number in Javascript

Question

I am trying to create the 3 digit random number using the following function:

gridView.generateData = function( size ) {

        var newData = [];

        var size = size || 100;
        var baseSize = 0;

        while( baseSize < size ) {
            baseSize++;
            newData.push( {
                "mk" : Math.floor( Math.random()*999 ) + 100 //3 digit
            } );
        }       
        return $q.all( newData );
    }

But in the out put object i see there is number of instance having 4 digit numbers to. how to insure only 3 digit using the above function?

thanks in advance.

Answer 1

if you want to generate 3 digit random numbers between 100 - 999 then you can try the below solution

Math.floor(Math.random()*(999-100+1)+100);

Answer 2

The problem in your code is that you are using +100 in your number, it will give 4-digits numbers for all generated numbers that are greater than 900: 900+100=1000.

Solution:

If you want to ensure that your number is always a 3-digits number(>=100), you can do it like this:

var number;
do {
  number = Math.floor(Math.random() * 999);
} while (number < 100);

You should just test upon this number until it's >=100, instead of adding 100 to it because it can give you numbers with 4 digits.

Demo:

Here's a live Demo which allows only numbers between 100 and 999:

var number;
do {
  number = Math.floor(Math.random() * 999);
} while (number < 100);
console.log(number);

Answer 3

You could use a variable for the digits. Then build the 10 potency otu of the digits and use this number as factor and as value to add for getting the digits with slice from the right side.

Later, you could return an obejct with a numerical value or some other formatting of the result.

function generateData(size) {
    var digits = 3;
    return Array.apply(null, { length: size || 100 }).map(function () {
        return Math.floor(Math.random() * Math.pow(10, digits) + Math.pow(10, digits)).toString().slice(-digits);
    });
}

console.log(generateData());

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

I had to face myself almost the same problem (generating "something" in the range from "000" to "999") and decided to use:

var round = 10;
while( round > 0 ) {
    round--;
    console.log( Math.random().toString().concat("0".repeat(3)).substr(2,3) );
}                

Should remember that Math.random() returns a random number between 0 (inclusive), and 1 (exclusive). You may get "0.001" or even "0.0001" (that is lot of leading zeros on the decimal part), so multiply by 1000 or 999 would not lead to the result I desired.

Answer 5

Don't add 100. Otherwise you will overflow.

Answer 6

You were on the right path, but that +100 was not necessary, as shown below:

gridView.generateData = function( size ) {
    var newData = [];

    var size = size || 100;
    var baseSize = 0;
    var digit = ~~( Math.random()*999 );

    while(digit < 100) {
        digit = ~~( Math.random()*999 );
    }

    if(digit >= 100) {
        while( baseSize < size ) {
            baseSize++;
            newData.push( {
                "mk" : digit //3 digits
            });
        }

        return $q.all( newData );
    }
}

As Math.random() returns a float between 0 and 1, so just multiply it by 999 without the +100.

Answer 7

Your code problem

if you add +100.its append 4 digit number .If you remove the 100. its append below 3 digit number

So better use like this, parseInt(Math.floor(math.random()*100000).toString().substring(0,3)) .Generate random num with above 4 digit, then cut the string with 3digit. And finally convert into integer .In this method they prevent the below and above3 digit value.

var newData = [];

        var size =100;
        var baseSize = 0;

        while( baseSize < size ) {
            baseSize++;
            newData.push( {
                "mk" : parseInt(Math.floor( Math.random()*100000).toString().substring(0,3))
            } );
        }       
      console.log( newData );

Answer 8

var number = 0;
var digits = '';

for (i = 0; i < 3; i++) {
    number = Math.floor(Math.random() * 10);
    digits = digits + number.toString();
}

Answer 9

The accepted answer doesn't allow for a three digit number below 100 (ie. 001).

This oneliner will return from 000 to 999.

var rand='',d=0; while(d<3){rand+=Math.floor(Math.random()*10);d++}