How to sort A-a-Z-z a list of names?

Question

I need to sort a list of data by alphabetical order but in A-a-Z-z, were the name Antony comes before the name antony and Zelda comes before zelda so the list looks like this:

- Abigail
- Antony
- abigail
- antony
- Zelda
- zelda

The basic

list.sort(function (a, b) {
     if (a.name > b.name) return -1;
     if (a.name < b.name) return 1;
     return 0;
});

is producing a list like this:

• Abigail
• Antony
• Zelda
• abigail
• antony
• zelda

preferable language: Javascript

Answer 1

There is a built-in for that:

let list = ["abigail", "Antony", "Abigail", "antony", "Zelda", "zelda"];

list.sort((a, b) =>
  a.localeCompare(b, "en", { caseFirst: "upper" })
);

console.log(list);

EDIT: maybe you want this?

let list = ["abigail", "Antony", "Abigail", "antony", "Zelda", "zelda"];

const compareUpperFirst = (a, b) => {
  if (a === "" && bb === "") return 0;
  if (a === "") return -1;
  if (b === "") return 1;
  let aa = a.charAt(0);
  let aal = aa.toLowerCase();
  let bb = b.charAt(0);
  let bbl = bb.toLowerCase();
  if (aal < bbl) return -1;
  if (aal > bbl) return 1;
  if (aa < bb) return -1;
  if (aa > bb) return 1;
  return compareUpperFirst(a.substr(1), b.substr(1));
};

list.sort(compareUpperFirst);

console.log(list);