javascript sort method not running on array[1]

Question

var test = ['hello', 'Hello']
var arg1 = test[0].split('').sort().join('').toLowerCase();  // ehllo
var arg2 = test[1].split('').sort().join('').toLowerCase();  // hello

Would someone be able to explain why the sort method appears to have no effect on the 2nd element of the test array?

`test[1].toLowerCase().split('').sort().join('');` should work. You are sorting then transforming. — Hanlet Escaño, Jul 14 '16 at 18:51
using toLowerCase() before the sort did the trick. Thank you for the great replies! Sorry I didn't find the previous thread on the subject. — adam.shaleen, Jul 14 '16 at 21:51

score 8 · Answer 1 · answered Jul 14 '16 at 18:50

console.log('H' < 'e', 'H'.charCodeAt(0), 'e'.charCodeAt(0));

Capital letters range from 65 to 90. Lower case letters range from 97 to 122. String comparisons are based on character codes.

Consider being more explicit with your sort by using a custom function.

score 4 · Accepted Answer · answered Jul 14 '16 at 18:52

4

If you want to ignore case, you should lowercase before sorting:

console.log('Hello'.toLowerCase().split('').sort().join('')); // ehllo

answered Jul 14 '16 at 18:52

Oriol

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score 2 · Answer 3 · edited May 23 '17 at 12:14

2

This question was answered in StackOverflow before case insensitive sorting in Javascript

'Hello'.split('').sort(function (a, b) {
  return a.toLowerCase().localeCompare(b.toLowerCase());
}).join('');

If you care to have a case sensitive output e.g. Hello would become eHllo

edited May 23 '17 at 12:14

Community

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answered Jul 14 '16 at 18:57

anvk

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javascript sort method not running on array[1]

3 Answers3