Sorting a string array with uppercase first including accents

Question

So I was wandering around the internet searching for some sorting function in js. Here is the problem. We have a string array like this :

['único', 'UNICO', 'árbol', 'ARBOL', 'cosas', 'COSAS', 'fútbol', 'FUTBOL']

and we want somthing like this (Uppercase first):

['ARBOL', 'COSAS', 'FUTBOL', 'UNICO', 'árbol', 'cosas', 'fútbol', 'único']

or like this (lowercase first):

['árbol', 'cosas', 'fútbol', 'único', 'ARBOL', 'COSAS', 'FUTBOL', 'UNICO']

The thing is : it's really easy to get this :

['ARBOL', 'COSAS', 'FUTBOL', 'UNICO', 'cosas', 'fútbol', 'árbol','único']

with the .sort(); function but we don't want the accentuated words at the end so we use the

.sort(function(a, b) {
  return a.localCompare(b);
});

but we end up with this ...

['ARBOL', 'árbol', 'COSAS', 'cosas', 'FUTBOL', 'fútbol', 'UNICO', 'único']

Do you guys have any idea on how to combine both ?

Answer 1

You can simply extend the sort function to prioritize uppercase characters in the beginning of strings:

const arr = ['ÁRBOL', 'único', 'UNICO', 'árbol', 'ARBOL', 'cosas', 'COSAS', 'fútbol', 'FUTBOL'];

function startsWithUppercase(str) {
    return str.substr(0, 1).match(/[A-Z\u00C0-\u00DC]/);
}

arr.sort(function(a, b) {
    if (startsWithUppercase(a) && !startsWithUppercase(b)) {
        return -1;
    } else if (startsWithUppercase(b) && !startsWithUppercase(a)) {
        return 1;
    }
    return a.localeCompare(b);
});

console.log(arr);