JavaScript's sort method handling of capital letters

Question

Noticed something potentially odd with JavaScript's sort() method. Given the following array:

var arr = ['Aaa',
'CUSTREF',
'Copy a template',
'Copy of Statementsmm',
'Copy1 of Default Email Template',
'Copy11',
'Cust',
'Statements',
'zzzz'];

Calling sort on this array:

console.log(arr.sort());

Yields:

["Aaa", "CUSTREF", "Copy a template", "Copy of Statementsmm", "Copy1 of Default Email Template", "Copy11", "Cust", "Statements", "zzzz"]

Is this correct? ie. CUSTREF is listed first, is this because of it's capital letters?

That's how [`sort()`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort) works. Characters are sorted by their Unicode values. — Teemu, Feb 11 '14 at 11:41
possible duplicate of [How to perform case insensitive sorting in Javascript?](http://stackoverflow.com/questions/8996963/how-to-perform-case-insensitive-sorting-in-javascript) — Liam, Feb 11 '14 at 11:43

score 20 · Accepted Answer · answered Feb 11 '14 at 11:41

20

That is correct. The strings are being sorted in a binary fashion, using the ordinal values of the characters themselves.

For a case-insensitive sort, try this:

arr.sort(function(a,b) {
    a = a.toLowerCase();
    b = b.toLowerCase();
    if( a == b) return 0;
    return a < b ? -1 : 1;
});

answered Feb 11 '14 at 11:41

Niet the Dark Absol

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Okay... which two chumps downvoted this answer just now, and why? – Niet the Dark Absol Feb 20 '14 at 14:35

Mika Tuupola · Answer 2 · 2014-02-11T11:54:18.463

13

You are correct, it is because of capital letters. If you are sorting strings which might have non ASCII characters such as ä and ö you should use String.localeCompare(). This also fixes the capital letter problem.

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

edited Feb 11 '14 at 11:54

answered Feb 11 '14 at 11:47

Mika Tuupola

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Ori Refael · Answer 3 · 2014-02-11T14:47:33.967

3

Yes, It has a higher Unicode value. (it = the 'U' in the first word)

you maybetter use

.sort(function(a,b) { return (a.toLowerCase() < b.toLowerCase()) ? -1 : 1;});

edited Feb 11 '14 at 14:47

answered Feb 11 '14 at 11:41

Ori Refael

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WRONG. The sort function must return a *number*, not a boolean. – Niet the Dark Absol Feb 11 '14 at 11:42
1

+ missing arguments `a` and `b`. – Teemu Feb 11 '14 at 12:04
@Teemu Pfft, that too. The boolean return was so blinding to me that I completely missed the missing arguments! XD – Niet the Dark Absol Feb 11 '14 at 12:38

score 0 · Answer 4 · answered Feb 11 '14 at 11:41

0

if you look at they way characters are encoded (eg. ASCII table) you will see, that capital letters have lower values that lowercase one's, so yes - it's because of capital letters

answered Feb 11 '14 at 11:41

pwolaq

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Sharanya Dutta · Answer 5 · 2014-02-11T12:09:51.540

0

Since U (U+0055) has a lesser Unicode value than o (U+006F), a case-sensitive sort will always place U before o. For a case-insensitive sort, you should try:

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

edited Feb 11 '14 at 12:09

answered Feb 11 '14 at 11:41

Sharanya Dutta

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JavaScript's sort method handling of capital letters

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