How to sort in javascript

Question

If I have an array like this:

var tab = ['1185 Design','3 D Exhibits','44Doors', '4Concepts','ABC Data','acceleration'];

And I want to sort it so that small letter 'a' element comes before capital letter 'A' element.

Answer 1

Use Array#sort() method with String#localeCompare()

var tab = ['1185 Design', '3 D Exhibits', 'nb', 'N', 'cd', '44Doors', '4Concepts', 'ABC Data', 'acceleration'];

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

function sortFn(a, b) {
  // if both are equal return 0
  if (a == b) return 0;
  // if first characters are equal call the same function with remaining (recursion)
  if (a.charAt(0) == b.charAt(0)) return sortFn(a.slice(1), b.slice(1))
  // check lowercase or uppercase based on that return value
  if (/^[a-z]/.test(a.charAt(0)) && /^[A-Z]/.test(b.charAt(0))) return -1;
  if (/^[a-z]/.test(b.charAt(0)) && /^[A-Z]/.test(a.charAt(0))) return 1;
  // otherwise ude normal compare function
  return a.localeCompare(b);
}

console.log(tab);

---

UPDATE : In case if you want to sort with alphabetical order and small letter should have higher precedence only if they are equal then do something like.

var tab = ['1185 Design', '3 D Exhibits', 'nb', 'N', 'cd', '44Doors', '4Concepts', 'ABC Data', 'acceleration'];

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

function sortFn(a, b) {
  // if both are equal return 0
  if (a == b) return 0;
  // if first characters are equal call the same function with remaining (recursion)
  if (a.charAt(0) == b.charAt(0)) return sortFn(a.slice(1), b.slice(1))
  // check lowercase or uppercasebased on that return value in case the letters are equal 
  if (a.charAt(0).toLowerCase() == b.charAt(0).toLowerCase()) {
    if (/^[a-z]/.test(a.charAt(0)) && /^[A-Z]/.test(b.charAt(0))) return -1;
    if (/^[a-z]/.test(b.charAt(0)) && /^[A-Z]/.test(a.charAt(0))) return 1;
  }
  // otherwise ude normal compare function
  return a.localeCompare(b);
}


console.log(tab);