JS sort() empty to end

Question

I have a JS sort like this:

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

This works, but some of my records are "" or null.

The empty records are listed at the begin but I want them at the end.

I think there are better ways to do this than:

if (a == "") a = "zzzz";

But how can I do this?

http://stackoverflow.com/questions/29829205/sort-an-array-so-that-null-values-always-come-last — Ofir Baruch, Oct 08 '15 at 12:45

Eric Hotinger · Accepted Answer · 2016-05-18T14:47:52.370

39

Maybe something like this:

records.sort(function(a, b) {
    if(a === "" || a === null) return 1;
    if(b === "" || b === null) return -1;
    if(a === b) return 0;
    return a < b ? -1 : 1;
});

edited May 18 '16 at 14:47

answered Oct 08 '15 at 12:44

Eric Hotinger

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2

care to explain with a diagram ? – Usman Iqbal Oct 08 '20 at 06:46

score 3 · Answer 2 · answered Sep 23 '21 at 22:10

3

records.sort((a,b) => a ? b ? a.localeCompare(b) : -1 : 1);

A ES6 one-liner version of Eric Hotinger's answer

answered Sep 23 '21 at 22:10

Joey Carlisle

586
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12

score 2 · Answer 3 · edited Apr 19 '19 at 10:02

2

Shorter way:

['a', '!', 'z','Z', null, undefined, '']
.sort((a, b) => !a ? 1 : (!b ? -1 : (a.localeCompare(b))) )

edited Apr 19 '19 at 10:02

Nino Filiu

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answered Apr 19 '19 at 08:42

Serhii Kimlyk

87
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JS sort() empty to end

3 Answers3

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