I have arrays such as
var arrayVal_Int = ["21", "53", "92", "79"];
var arrayVal_Alpha = ["John", "Christine", "Lucy"];
var arrayVal_AlphaNumeric = ["CT504", "AP308", "NK675"];
arrayVal_Int should be considered as (purely) numeric.arrayVal_Alpha and arrayVal_AlphaNumeric should be considered as strings.I need to check that in JavaScript.
Shortest solution, evals to true if and only if every item is (coercible to) a number:
!yourArray.some(isNaN)
"" that will render as false on isNaNnull coerces to zero. If you need null to make the code return false, you need to replace isNaN with something more specific.I had a similar need but wanted to verify if a list contained only integers (i.e., no decimals). Based on the above answers here's a way to do that, which I posting in case anyone needs a similar check.
Thanks @Touffy, for your suggestion.
let x = [123, 234, 345];
let y = [123, 'invalid', 345];
let z = [123, 234.5, 345];
!x.some(i => !Number.isInteger(i)) // true
!y.some(i => !Number.isInteger(i)) // false
!z.some(i => !Number.isInteger(i)) // false
x.every(Number.isInteger) instead of the double negation :)every if there was a native isNumber function. Since isNaN is the opposite of isNumber and I wanted to use it without creating an extra function isNumber = n => !isNaN(n), I exploited the fact that ∀𝑛|¬𝜆(𝑛) is logically equivalent to ∄𝑛|𝜆(𝑛). But you don't have to do that with isInteger.Using simple JavaScript, you can do something like this:
var IsNumericString = ["21","53","92","79"].filter(function(i){
return isNaN(i);
}).length > 0;
It will return true;
Try this:
let x = [1,3,46,7,7,8];
let y = [1,354,"fg",4];
let z = [1, 3, 4, 5, "3"];
isNaN(x.join("")) // false
isNaN(y.join("")) // true
isNaN(z.join("")) // false
"1", "e5"), and vice-versa ("1.2", ".3").