Identify if all characters are equal

Question

I have the following code that works perfectly.

In it I possess a string and check whether all of its characters are equal or not:

var numbers = '1111121111',
    firstNumber = numbers.substr(0,1),
    numbersEquals = true;

for(let i = 1; i < numbers.length; i++) {
  if (numbers[i] != firstNumber) numbersEquals = false;
}
console.log('numbersEquals = ' + numbersEquals)

There is an easier way or a ready method to do this?

I think I’m using too much code to do something simple.

Answer 1

With ECMA6, it is possible using Set:

const unicos = [...new Set(numbers)]

Example:

var numbers = '1111112111';

if (new Set(numbers).size > 1)
  console.log("Todos os valores não são iguais");
else
  console.log("Todos os valores são iguais");

Edit:

As @bfavaretto suggested, it is possible to get the number element in Set object.

Answer 2

There are many ways to do this. I don’t think yours is wrong, it’s very clear and didactic. But it has shorter ways, like Mr Felix’s. Converting to array can do some tricks too, for example:

var numbers = '1111121111';
numbers.split('').every(function(num, i, arr) { return num == arr[0] }); // false

Or even shorter in ES-2015:

const numbers = '1111121111';
[...numbers].every( (num, i, arr) => num == arr[0] ); // false

Answer 3

Using RegExp.prototype.test(), do the following:

/^(.)\1+$/.test(sua_string)

This returns true or false depending on whether all characters are equal or not.

/^(.)\1+$/.test("xxxx"); // true
/^(.)\1+$/.test("xxxy"); // false

test() is since the ECMA 3rd Edition (1999). It will basically work on all browsers.

Answer 4

I would do exactly what you did because it should be the fastest solution. But there are several alternatives, one of them:

var numbers = '1111121111';
console.log('numbersEquals = ' + (numbers.replace(new RegExp(numbers.substr(0, 1), 'g'), "").length == 0));
numbers = '1111111111';
console.log('numbersEquals = ' + (numbers.replace(new RegExp(numbers.substr(0, 1), 'g'), "").length == 0));

I put in the Github for future reference.