ESLint Unexpected use of isNaN

Question

As the documentation suggests, use Number.isNaN.

const isNumber = value => !Number.isNaN(Number(value));

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Quoting Airbnb's documentation:

Why? The global isNaN coerces non-numbers to numbers, returning true for anything that coerces to NaN. If this behavior is desired, make it explicit.

// bad
isNaN('1.2'); // false
isNaN('1.2.3'); // true

// good
Number.isNaN('1.2.3'); // false
Number.isNaN(Number('1.2.3')); // true

FYI, this will not work for IE. Check here at browser compatibility.

@Andy Gaskell isNumber('1.2.3') return true, you might want to edit your answer and use Number() in place of parseFloat()

    const isEmpty = value => typeof value === 'undefined' || value === null || value === false;
    const isNumeric = value => !isEmpty(value) && !Number.isNaN(Number(value));

  console.log(isNumeric('5')); // true
  console.log(isNumeric('-5')); // true
  console.log(isNumeric('5.5')); // true
  console.log(isNumeric('5.5.5')); // false
  console.log(isNumeric(null)); // false
  console.log(isNumeric(undefined)); // false

In my case, I wanted to treat 5 (integer), 5.4(decimal), '5', '5.4' as numbers but nothing else for example.

If you have the similar requirements, below may work better:

const isNum = num => /^\d+$/.test(num) || /^\d+\.\d+$/.test(num);

//Check your variable if it is a number.
let myNum = 5;
console.log(isNum(myNum))

To include negative numbers:

const isNum = num => /^-?\d+$/.test(num) || /^-?\d+\.\d+$/.test(num);

This will remove your issue of global use of isNaN as well. If you convert the isNum function to a normal ES5 function, it will work on IE browser as well.